Google
This is a digital copy of a book lhal w;ls preserved for general ions on library shelves before il was carefully scanned by Google as pari of a project
to make the world's books discoverable online.
Il has survived long enough for the copyright to expire and the book to enter the public domain. A public domain book is one thai was never subject
to copy right or whose legal copyright term has expired. Whether a book is in the public domain may vary country to country. Public domain books
are our gateways to the past, representing a wealth of history, culture and knowledge that's often dillicull lo discover.
Marks, notations and other marginalia present in the original volume will appear in this file  a reminder of this book's long journey from the
publisher lo a library and linally lo you.
Usage guidelines
Google is proud lo partner with libraries lo digili/e public domain materials and make them widely accessible. Public domain books belong to the
public and we are merely their custodians. Nevertheless, this work is expensive, so in order lo keep providing this resource, we have taken steps to
prevent abuse by commercial panics, including placing Icchnical restrictions on automated querying.
We also ask that you:
+ Make n on commercial use of the files We designed Google Book Search for use by individuals, and we request thai you use these files for
personal, non commercial purposes.
+ Refrain from automated querying Do not send automated queries of any sort lo Google's system: If you are conducting research on machine
translation, optical character recognition or other areas where access to a large amount of text is helpful, please contact us. We encourage the
use of public domain materials for these purposes and may be able to help.
+ Maintain attribution The Google "watermark" you see on each lile is essential for informing people about this project and helping them find
additional materials through Google Book Search. Please do not remove it.
+ Keep it legal Whatever your use. remember that you are responsible for ensuring that what you are doing is legal. Do not assume that just
because we believe a book is in the public domain for users in the United States, that the work is also in the public domain for users in other
countries. Whether a book is slill in copyright varies from country lo country, and we can'l offer guidance on whether any specific use of
any specific book is allowed. Please do not assume that a book's appearance in Google Book Search means it can be used in any manner
anywhere in the world. Copyright infringement liability can be quite severe.
About Google Book Search
Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers
discover the world's books while helping authors and publishers reach new audiences. You can search through I lie lull lexl of 1 1 us book on I lie web
al _.:. :..:: / / books . qooqle . com/
IRST BOOK
in CEF1ERAL
HATH E H ATI CS
ATUCbb
^
L
^A^urr io<v q. n°
HARVARD COLLEGE
LIBRARY
GIFT OF THE
GRADUATE SCHOOL
OF EDUCATION
3 2044 096 990 197
FIRST BOOK
IN GENERAL
MATHEMATICS
BY
ROBERT KING ATWELL
HEAD OF DEPARTMENT OF MATHEMATICS,
UNIVERSITY OF PORTO BICO
DIAGRAMS BY
FRANK S. PUGH
.BUPEBVIBOB OF MECHANIC ABT8,
DBPABTMENT OF EDUCATION, PORTO RICO
NEW YORK
PARKER P. SIMMONS CO., Inc.
^LdiJL^ t _v ^ihixin °
L • : v.', : THE
oh/vLrL 4 it .:.;• JJL OF F0U£ATI(Wf
lAc «.<
^ t \ f\ TL<v
Copyright, 1917, s
BT
I
• PARKER P. SIMMONS CO., INC
j .
CONTENTS
CHAPTER I Page 1
» * * * » » * .
Lines and> Angles
The straight line: plane: right angle: straight angle: cirfcle:
degrees, minutes and seconds: change of angular units: addition
and subtraction of angles.
I. Shop Exercise: leveling devices and appurtenances.
CHAPTER II Page 14
' Straight Angles and Perpendiculars
Axioms: supplementary angles: vertical angles: postulates:
theorems concerning perpendiculars: discussion of the straight
line.
II. Field Exercise: location of corner stakes for exca
vation for tpe foundation of a building.
CHAPTER III . . .• . . . . . Page 30
Parallel Lines
,. Perpendiculars to same line: parallels to a given line: aline
perpendicular to one of two parallels: alternate interior angles:
exteriorinterior angles.
Ilia. Shop Exercise: parallel, ruler.
Illb. Field Exercise: marking corners by batter boards.
Completion of work of laying out ground for excavation,
begun in Field Exercise II.
• • •
111
iv CONTENTS
CHAPTER IV Page 41
Tkiangles
Theorems for proving two triangles equal: isosceles triangle:
bisector of given angle: construction of triangle when three
sides are given: sum of the angles of a triangle: exterior angle
of a triangle.
IVa. Class Room Exercise: combining angles of paper
TRIANGLES.
IVb. Field Exercise: measuring through an obstruction.
CHAPTER V Page 52
Circles
Theorems involving equal arcs, equal central angles, equal
chords: construction of an angle equal to a given angle: con
struction of a circumference through three given points not in
a straight line: measurement of an inscribed angle.
V. Field Exercise: laying out a road in an arc op a
CIRCLE PASSING THROUGH THREE GIVEN POINTS.
CHAPTER VI . . ... . . . Page 65
Polygons: Use op Letters to Express General Laws
Classes of polygons: proof of formula for the sum of the angles
of a polygon for particular polygons and for polygons in general,
with introduction of algebraic symbols: the equation: construc
tion of regular polygons.
VI. Shop Exercise: design op plate pad with ornamenta
tion BASED ON ABOVE CONSTRUCTION.
CHAPTER VII Page 77
Literal Expressions
Definitions : combination of terms : signs of inclosure : exponents :
like terms: addition and subtraction of monomials and of poly
nomials: removal of signs of enclosure.
Vila. Field Exercise: profile op land. •
Determination of the various heights and depressions along
a line.
Vllb. Drawing Exercise: a plot of above*
CONTENTS v
CHAPTER VIII ....... Page 90
The Construction and Application of Parallel Lines
Construction of a line through a given point parallel to a given
straight line: theorems on parallelograms: derivation of method
of dividing a given line into any number of equal parts: the
vernier.
Villa. Field Exercise: staking off parallels.
Laying out a path parallel to the front of a building, or
other prominent straight line.
VTIIb. Drawing Exercise: construction of vernier.
CHAPTER IX Page 102
Areas
Tables of measure : theorems concerning the area of the rectangle,
the parallelogram, the triangle, the trapezoid: application of
square root to equation.
IX. Field Exercise: finding area of a field by base and
ALTITUDE METHOD.
Division of the field into triangles, and measurement of
their bases and altitudes.
CHAPTER X Page 111
Multiplication of Literal Expressions; 'Factoring
Product of a minus quantity by a minus quantity: multiplica
tion of monomials, of polynomials: the square of the sum of two
quantities, the square of the difference of two quantities, the
product of the sum and the difference of two quantities (deriva
tion of these rules algebraically, with geometrical proof) : reversal
of these rules for factoring.
X. Shop Exercise: construction of transit.
Execution of directions in the accompanying working draw
ings.
CHAPTER XI Page 133
The Right Triangle
Complementary angles: the right triangle: detailed geometric
proof that the square of the hypotenuse of a right triangle
yi CONTENTS
equals the sum of the squares of the legs: quadratic surds:
projection: the square of the side of any oblique triangle.
XIa. Shop Exercise: construction of stairs.
Xlb. Shop Exercise: model of above.
CHAPTER XII Page 152
Literal Fractions: Heron's Formula *
Reduction of literal fractions to lowest terms: least common
multiple: least common denominator: combination of literal
fractions: changing mixed numbers to improper fractions:
fractional equations: algebraic derivations of formula for find
ing the area of a triangle.
Xlla. Shop Exercise: erection of stairs.
Location of stairs made in Shop Exercise XI.
XHb. Field Exercise: area of field by heron's formula.
Division of a field into triangles, and measurement of
their sides.
CHAPTER XIII . . . • , . . Page 167
Similarity
Similar triangles: proportion: measurement of inaccessible dis
tances by proportion.
XIII. Field Exercise: finding width of river.
Application of similar triangles and proportion.
CHAPTER XIV Page 177
Functions of Angles
Trigonometric ratios: functions of an angle: relation between
functions of an angle and its complement: table of functions
to every 5°: solution of right triangles.
XI Va. Shop Exercise: design of rafter.
Application of similar right triangles.
XlVb. Shop Exercise: design of flight of stairs for
ANY GIVEN LOCATION.
Application of similar right triangles.
CONTENTS vii
CHAPTER XV ... . . . . . Page 189
Computation of Functions: Four Place Table of
Sines and Cosines and Its Use
Computation of functions of 30°, 60°, and 45°: to find functions
by plotting: four place table of sines and cosines: to find the
sine of any given angle: to find the cosine of any given angle:
to find, the angle corresponding to any given sine: to find the
angle corresponding to any given cosine: solution and check
ing of compass surveys.
XV. Field Exercise: finding area of field by triangula
tion.
Dividing a field into triangles, measuring a base line and
the angles of the triangle.
CHAPTER XVI Page 207
Relations Between the Functions: Axes of Reference:
Functions of Obtuse Angles
Sine divided by cosine equals the tangent: sine squared plus
cosine squared equals unity: change of values of sine and cosine:
axes of reference: functions of an angle in the first quadrant,
and of an angle in the second quadrant: relation between the
functions of an angle and the functions of its supplement:
plotting change of values of sine and cosine: practical consider
ations in use of tables.
XVI. Field Exercise: measurements for topographical
MAP OF TRIANGULATION.
Location of topographical features in fields measured in
Field Exercise XV.
CHAPTER XVII Page 226
Solution of Oblique Triangles
Number of cases of solvable oblique triangles: Sine Law:
Cosine Law: application of preceding laws to solution of tri
angles: discussion of the accuracy of computations: proof of
theorems to aid in checking results: a side of a triangle divided
by the sine of its opposite angle is numerically equal to the
diameter of the circumscribed circle: area of triangles.
XVII. Drawing Exercise: map of survey.
Plot of measurements made in Field Exercises XV and XVI.
viii CONTENTS
CHAPTER XVIII Page 2S0
Simultaneous Equations: Finding Heights op Hills:
Ambiguous Case
Solution of simultaneous equations: application of simultaneous
equations to finding heights of hills: discussion of the ambiguous
case in the solution of triangles.
XVIII. Field Exercise: finding the height or a hill.
Application of simultaneous equations.
PREFACE
It has been urged that the investigation and progress
that have characterized other branches of the school cur
riculum have been lacking in so far as mathematics is con
cerned, especially in the case of mathematics as applied to
secondary schools. This is doubtless due largely to the fact
that mathematics, as a pure science, is not so susceptible to
theory as is a subject whose limitations are not so closely
drawn, and whose subject matter is more open to specula
tion. In spite of this fact, however, educators have dreamed
of a more ideal course in mathematics; a course which would
give better and larger returns for the time spent in study,
and a course which would remove from mathematics the
stigma which it so often bears, of being the bite noir of the
average high school student.
# The teacher of secondary mathematics has his choice
between what might be called the natural and the artificial
incentives. Under the natural incentives fall the following:
a — the uses of mathematics in the activities of life; b — the
charm of achievement which comes with the solving of
problems; c — the gain of mental power, of the ability to rea
son clearly to a definite conclusion. Among the artificial
incentives, the following are the most usual and most power
ful: a — graduation from the high school; b— preparation
for college; c— the winning of some special prize or honor;
d — the avoiding of suspicion of mental weakness.
The average teacher is content to use one or more of
the artificial incentives, largely because it is easier to do so
than to exert the necessary personality which is required ta
ix
X PREFACE
use successfully the natural incentives. Moreover, most
secondary school textbooks in mathematics seem to have
been written with the purpose of requiring students to cover
a certain amount of work, rather than to stimulate interest
in the work being done.
The pedagogical remedy for such a state of affairs may
be summed up in the words "practical use" or " relation to
life." Interest can be aroused and maintained only when
there is a tangible relation between the subject matter
studied and the " outofschool " life of the student. This
idea is not new, but has been applied in various forms at
different times. In the Colebrook School, for example, in
the northern part of New Hampshire, where the school is
being made over to fit the needs of the community, the
course in geometry has been substituted by a course in
practical mathematics which includes: a — the algebra of
the equation; 6 — the application of geometry to practical
measurements ; c — the elementary principles of surveying.
Professor Parker, in his book on High School Methods,
has the following to say regarding highschool mathematics :
"The order of topics in a subject should be decided by the
needs, capacities and interests of highschool students, not
by the nature of the subject itself, nor the interests of a
specialist in the subject. In highschool mathematics this
standard will require the inclusion of many practical prob
lems and the mixing of the easier topics of algebra and
geometry, and some arithmetic, in the first year."
Mathematicians have often claimed that we should have
texts covering a course in "general mathematics," rather
than break up our course into algebra, geometry and trig
onometry. Attempts to introduce a course which would
cover these three phases of mathematics at the same time,
KtEPACE xi
have liot proven generally successful, due largely to a lack of
pedagogical and psychological principles in the working out
of the content of the book.
In view of these facts the author of this work has felt
that there is a place in the intermediate high school, where
a course that will sum up the arithmetic of the grades, and
give some insight into the coming mathematical problems
of algebra and geometry, might be of use, especially if this
course is linked up with the work in mechanical drawing,
manual training and elementary surveying, so that the stu
dent may see the application of the mathematical theories
that he is studying, and have the pleasure of actually accom
plishingr'something in the way of a simple surveying instru
ment and a map of the locality. There is enough of the
explorer in the blood of the average youth to lend romance
to the prosaic task of land measurement, and this may be
counted on to stimulate and maintain interest.
Another place which this work will fill is that of a review
course in mathematics for prospective teachers. Normal
students who have finished their highschool mathematics
ought to have a general course covering the most important
points of elementary and secondary mathematics, worked
out from the standpoint of a teacher, and with emphasis on
" application.' ' Too many prospective teachers have cov
ered their arithmetic, their algebra and their geometry,
simply because it was required of them, have mastered
mathematical principles, and never thought of applying
these principles to problems outside the schoolroom. Such
a course as that outlined in this book will compel the student
to apply theory to practice.
And finally, there is the satisfaction and advantage that
must be the reward of the man who knows something of
xii PREFACE
mensuration: who can, in ease of necessity, make his own
apparatus and lay out a path, a fence line, a ditch, or the
boundary of his property: who can see that his time spent
in school has had a certain practical, as well as social value;
and who is a better citizen, and a better neighbor, because
his range of vision has been broadened, his sympathies have
been extended, and instead of a theorist, he is a practical,
useful exponent of the modern ideas in education,
Frederick Kurtland Fleagle,
Dean, University of Porto Rico.
March 13, 1916.
FIRST BOOK IN
GENERAL MATHEMATICS
CHAPTER I
LINES AND ANGLES
1. A true Line has neither thickness nor width; it has
length only.
Note: A fine line drawn on mechanical drawing paper with a hard pencil
is not a true line, but is near enough to it for practical purposes.
2* The ends of a line are called its Extremities.
3. When one line lies completely in another line it is said
to Coincide with it.
4. When two straight lines cross each other, they are said
to Intersect.
5* The intersection of two straight lines is a Point.
6. A Point has no dimensions : only Location.
7a. A Plane is a flat surface, having only two dimensions —
Length and Breadth ; it is a surface such that a straight line
joining any two points in it lies entirely in the surface.
The surface of the blackboard may be considered a plane.
7b. A Plane Figure is a figure outlined in a Plane — it has
no thickness.
Plane figures are not actual objects — they are Abstract
Ideas. In forming an Abstract Idea, the mind abstracts
certain qualities of the object and considers them only.
Abstract ideas are at the basis of Arithmetic, as for
example: when it is said "two plus three is five," instead
l
2 GENERAL MATHEMATICS
of saying: "two apples plus three apples is five apples" or
"two books plus three books is five books/' etc. Here it is
the quality of number that is abstracted from the objects
added, and their other qualities such as color, or hardness,
etc., are not considered; and so we are enabled to state, a
general truth that may be* applied to all kinds of objects.
So, in Reasoning about Plane Figures, we are abstracting
the ideas of length and breadth, to derive truths which may
later be applied to all kinds of objects.
8. Properties of the Straight Line:
(a) Only one straight line can be drawn from one point
to another; hence, 
(b) Two points determine a straight line; hence,
(c) Two straight lines which have two points in common
coincide and form but one line; hence,
(d) Two straight lines are equal if their extremities coin
cide; also,
(e) Two straight lines can intersect in but one point
(for if they had two points in common they would coincide
and not intersect) .
(f) A straight line is the shortest distance from one point
to another.
(g) A straight line can be produced indefinitely; that is
if the length of a line is not definitely stated, it may be
extended to any length desired.
9, The amount of opening between two intersecting
straight lines is called an Angle. The two straight lines are
called the Sides of the angle, and the point of intersection, the
Vertex, (The plural of vertex is Vertices.)
5 The vertex of this angle is the point A
and the sides of this angle are the straight
q lines AB and AC. In reading the letters
n of the angle, the letter at the vertex is
FlG * L always read in the middle. Thus, this
angle is read "The Angle BAC."
GENERAL MATHEMATICS 3
When no other angle has the same vertex, the angle may
be designated by reading the letter at the vertex only.
Thus, this angle may be read "The angle A."
Figure 1 also shows the method of marking points and
straight lines — a point being designated by one letter, as
"the point A" and a straight line by two letters, as "the
line AB."
As the sides of an angle are straight lines, as much or as
little of them as desired may be laid off from the vertex
(see 8g); that is:
10. The size of an angle is independent of the length of its sides.
The sign Z means "Angle." Note that the lower line of
this sign is horizontal. A means "Angles."
11. Adjacent Angles are angles which
have a side in common (that is, be
longing to both of them) and the same
vertex.
Thus, Z CAB and Z BAD are adjacent FlQ  2 
angles. This may be written "CAB and BAD are adj. A"
12a. If one straight line which intersects a second straight
line divides the space on one side of the second straight line
in such a manner that the two adjacent angles thus formed
are equal, these two adjacent angles are called Right Angles
and the two lines are said to be Per c
pendicular to each other.
Thus CD is perpendicular to AB
at the point D and the angles CD A
and CDB are Right Angles. A per A
6
pendicular line may be called "a Q
perpendicular." Thus CD is a per Per
pendicular to AB. CD A and CDB are Right Angles.
12b. If one line is perpendicular to a second, then the
second is perpendicular to the first.
Thus, in Fig. 3 it may be said either that CD is perpen
dicular to A B or A B is perpendicular to CD.
4 GENERAL MATHEMATICS
13. The point of intersection of the perpendicular with
the other line is called the Foot of the perpendicular.
Thus, in Fig. 3, D is the Foot of the perpendicular CD.
The sign J. means "perpendicular."
Rt. / means "Right Angle." Rt. A means "Right Angles."
14. A Straight Angle is an angle whose sides extend in
opposite directions so as to form one straight line.
Thus ADB is a straight angle (Fig. 3).
15. A Right Angle is half a straight angle. (See art. 12
and 14.)
16. A Circle is a part of a plane bounded by a curved line,
called the Circumference of the circle, all points of which
are equally distant from a point within called the Center.
17. A straight line drawn from the center of the circle to
the circumference is called the Radius of the circle. (The
plural of Radius is Radii.)
18. All radii of the same circle are equal. (See art. 16.)
19. A straight line drawn through the center of the circle
and terminated at both ends by the circumference is called
a Diameter of the Circle.
20. A diameter is equal to two radii.
(See art. 17 and 18.)
21. An Arc is any part of the circum
A ference. Thus, in Fig 4, BA is an arc of
the circle whose center is C, the straight
line DA, passing through C, is a diameter,
FlQ  4  and DC, BC, and CA are Radii.
22. The angle formed between two Radii with its vertex
at the center is called a Central Angle.
Thus (Fig. 4) BCA is a central angle. DCB is also a
central angle.
23. The arc cut off by the Radii forming the sides of a
central angle is called its Intercepted Arc.
Thus (Fig. 4) BA is the arc intercepted by the central
angle BCA.
GENERAL MATHEMATICS 5
24. A circumference of a circle is divided into 360 parts,
called Degrees. A degree is subdivided into minutes and
seconds, according to the following table:
1 Circumference = 360 Degrees (°)
1 Degree = 60 Minutes (')
1 Minute = 60 Seconds (")
Note that the symbol for "degree" is a small circle written
above and to the right of the number of degrees, the symbol
for minutes is one small slanting line written above and to
the right of the number of minutes, and the symbol for
seconds is two small slanting lines written above and to
the right of the number of seconds. Thus, 17° 21' 30" is
read "seventeen degrees, twentyone minutes and thirty
seconds."
If Radii are drawn from the ends of each degree of arc to
the center of the circle, 360 small angles will be formed
with their vertices at the center of the circle. Now, since
any point in a plane may be taken as the center of a circle,
it follows that:
25. The sum of the angles about a point in a plane is equal
to 360 degrees.
Since each degree of arc has an angle of one degree at the
center of the circle it follows that:
26. A central angle is measured by its intercepted arc.
That is, there are just as many degrees, minutes and
seconds in a central angle as there are in its intercepted arc,
and so the above table serves to measure angles as well as
arcs.
Since the sides of a central angle, being straight lines,
may be produced indefinitely (8g), and become the Radii
of any larger circle, it follows that :
27. The size of an angular degree, or of any number of de
grees, minutes and seconds, is independent of the length of the
radii of the circle.
6 GENERAL MATHEMATICS
Exercise
. 1. Change 27° 12' 31" to seconds.
Since each degree contains 60 minutes we first multiply
27 by 60. To this product we then add the 12', obtaining
the total number of minutes. We then multiply this num
ber by 60, since there are 60" in each minute, and to this
number add the 31" and so obtain the total number of
seconds.
27° 1632'
60 60
1620' 97920"
12 31
1632' 97951" Ans.
2. Change 39° 28' 47" to seconds.
3. « 11° 0'51" «
4. " 46° 19' 41" «
6. « 67° 27' 49" «
6. * 210,608" to degrees, minutes and seconds.
60 ) 210,608" ( 3510' 8"
180 Since there are 60" in each minute
~30g we divide the total number of sec
300 onds by 60 to obtain the number of
minutes, and obtain 3510 for a quo
tient with a remainder of 8 — that
is, the given number of seconds is
8 equal to 3510' 8". We next divide
/ the number of minutes by 60 to
60 ) 3510 ( 58 30 obtain the number of degrees and
300 # so obtain the quotient 58 and the
510 remainder 30. The complete an
480 swer is therefore 58° 30' 8".
30
60
60
a a
u u u u
« u a u
u u u a
GENERAL MATHEMATICS
7. Change 26,821" to degrees, minutes and seconds.
8. « 46,429" « " *
9. « 37,922" «
10. « 67,908" "
11. « 189,647" «
12. Add 82° 47' 12", 21° 28' 21" and 42° 56' 29".
We first write the quantities in columns, keeping degrees
under degrees, minutes under minutes and seconds under
seconds, thus,
Adding the seconds we obtain the number
82° 47' 12" 62; from this we subtract 60", and add 1 to the
21° 28' 21" number of minutes. This leaves 2 as the
42° 56' 29" number of seconds. Adding the number of
147° 12' 2" minutes (not forgetting to add the 1 minute
carried from the seconds column), we obtain
132 as the number of minutes. Prom this subtract 120
(the number of minutes in two degrees) and we have 12 as
the number of minutes, with 2 to be added to the number
of degrees. Adding the degrees, we obtain 147, and have as
the complete answer 147° 12' 2".
Prom this example the student may understand the fol
lowing:
28. Rule : To add angles, write them one under the other,
keeping the degrees, minutes and seconds in separate columns;
add the seconds column, and if the number of seconds so obtained
is large enough to contain 60 or any multiple of it, subtract the 60
or its multiple so contained, from the number of seconds, leaving
the remainder to be the number of seconds in the .answer; change
the number of seconds subtracted to minutes, to be added to the
minutes column: add the minutes column (including the min
utes carried from the seconds column) and if the number of
minutes thus obtained contains 60 or any multiple of 60,
subtract the 60 or its multiple so contained from the number of
minutes, leaving the remainder to be the number of minutes
in the answer; change the number of minutes subtracted to
8 GENERAL MATHEMATICS
degrees to carry them to the degrees column and add the degrees
column. The complete answer is the sum of the quantities writ
ten under the columns of degrees, of minutes, and of seconds.
Add:
U 190° 17' 11' l*. 18° 39' 42' «. 26° 41 '17'
26° 24' 24' 31° 32' 26' 32° 22' 43'
17° 8' 16' 42° 14' 19' 18° 26' 57'
37° 31 '19'
i«. 27° 39' 42' 17. 42° 18' 36' . l*. 61° 18' 48'
36° 43' 16' 17° 11 ' 47' 32° 17' 39'
52° 16' 17' 29° 16' 29' 41° 18' 17'
61° 17' 38' 36° 14' 42' 27° 25' 35'
72° 38' 51' 33° 18' 15'
32° 27' 49' 62° 31' 56'
it. 24° 17' 43' ». 11° 39' 02'
108° 36' 29' 24° 36' 42'
0°10'51' 28° 0'05'
10° 46' 32' 42° 27' 43'
7° 0'41' 28° 49' 07'
5° 11' 17' 16° 20' 00'
2L Subtract 17° 54' 32' from 42° 19' 28'.
We first write the subtrahend beneath the
42 19 28 minuend, keeping the degrees, minutes and
17 54 32 seconds in separate columns. Since 28 is not
24° 24' 56' large enough for 32 to be subtracted from
it, we bring over 1' from the 19', leaving 18 as
the number of minutes in the minuend, and add the 1',
changed to 60', to the seconds in the minuend, making 88
as the number of seconds in the minuend. Subtracting 32
from 88 we have 56 as the number of seconds in the answer.
In the minutes column (since 54 cannot be subtracted from
GENERAL MATHEMATICS 9
18) we borrow 1 from the number of degrees in the minuend
and changing it to 60 minutes, we add it to the 18, making
78' the number of minutes in the minuend. Subtracting
we have 24' as the number of minutes in the answer. Then
subtracting 17° from 41° (the number of degrees left in the
minuend), we have 24° as the remainder, making 24° 24' 56
as the complete answer. From this example we may under
stand the following :
29. Rule : To subtract one angle from another, write the
subtrahend beneath ' the minuend, keeping degrees, minutes
and seconds in separate columns. If the number of seconds
in the minuend is less than that in the subtrahend, bring over
one minute from the minutes column in the minuend and add
it as 60 seconds to the seconds in the minuend. Subtract
the seconds in the subtrahend from the seconds in the minuend,
writing the difference as the number of seconds in the answer;
repeat this process for the minutes column, writing the difference
as the minutes in the answer, and finally subtract the degrees
in the subtrahend from those in the minuend, writing the
difference as the number of degrees in the answer.
Do the following examples in subtraction, not forgetting
that when it is necessary to bring over from the minutes
column before subtracting the seconds, the number of min
utes is thereby diminished by one, and that if it is necessary
to borrow from the degrees column before subtracting the
minutes, the number of degrees is thereby diminished by one.
22. 41° 18' 42" 23. 28° 17' 6" 24. 36° 17' 18"
27° 12' 37" 12° 11' 19" 29° 32' 26"
26.
66° 14' 8"
42° 37' 29"
28.
36° 17' 58"
9° 18' 10"
26.
112°
0'
16"
89°
42'
56"
29.
189°
0'
12"
72°
3'
15"
27.
79° 0' 0"
42° 18' 42"
30.
47° 10' 5"
46° 10' 26"
10
GENERAL MATHEMATICS
Shop Exercise
la. Though simple in construction, the Water Level
shown in the accompanying illustration gives very accurate
results. The pieces should be first nailed together and the
implement placed on some smooth surface, — as a table top.
Then, with a sharp pencil, the centers for the holes should
*HJ<
VAT&t L£V£L
be marked, great care being taken that they are both the
same vertical distance above the table top. The boring of
these holes completes the instrument, and when it is floated
in a pail nearly filled with water, the line of sight obtained
by looking through the holes is a true level line.
lb. The Ranging Pole is a pole about six feet long
and about one and onehalf inches in diameter. It should
be rounded, or at least the corners should be planed off, for
greater ease in handling.
GENERAL MATHEMATICS 13
The Ranging Pole should be brightly painted — preferably
in alternate bands of red and white — the bands being about
a foot wide. Two of these ranging poles are needed by a
surveying section.
Ic. The Leveling Rod is really a long ruler. It should
be made of a strip of straight, wellseasoned wood, at least
six feet long. The strip should first be painted white.
When it is dry, black stripes should be painted . entirely
across the strip at the end of each foot, and the number of
feet written just above these strips. Black marks reaching
halfway across the strip should be painted at the end of each
six inches. Short marks may be made at the end of each
inch. The bottom of the rod should be protected with a
strip of tin or other metal.
Id. Farm Level: • Work may be begun at this time
on the Farm Level, as illustrated in the accompanying dia
grams, which are selfexplanatory. Since the completion
of this instrument requires more time than it may be possible
to spend in the shop before completing the next few chapters,
it is suggested that the simple Water Level be made first,
for use in the Field Exercises immediately following.
The "telescope/' so called, in the diagram requires no
lenses — it derives its name from the surprising clarity of
vision obtained by looking through two holes at opposite
ends of an otherwise darkened tube.
CHAPTER n
STRAIGHT ANGLES AND PERPENDICULARS
30. An Axiom is a selfevident truth.
For example, the properties of a straight line are axioms.
31. Axiom: Quantities equal to the same quantity or to
equal quantities are equal to each other.
32a. Axiom: The whole is equal to the sum of all its
parts.
32b. Axiom : If equal subtrahends are subtracted from equal
minuends, the remainders are equal.
33a. Axiom : The whole is greater than any of its parts.
33b. Axiom: The halves of equals are equal.
34a. Axiom : The doubles of equals are equal.
34b. Axiom: If equals are added to equals the sums are equal
If two plane angles are equal, one may be thought of as
being placed on another so their vertices coincide and their
sides coincide, that is:
35a. Axiom: If two plane angles are equal they may be
made to coincide throughout.
35b. Axiom: If two plane angles may be made to coincide
throughout they are equal.
Note: In the following proof the two angles are proved equal by placing
the first angle on the second, so that their vertices coincide, and one side of
the first coincides with one side of the second. (This much might be done
with any two angles, whether they were equal or not.) Then it is necessary to
prove the second side of the first angle coincides with the second side of the
second angle, that is, that they coincide throughout.
/. is the symbol for "therefore."
> is the symbol for "is greater than."
14
GENERAL MATHEMATICS 15
Thus "AB > CD" is read "AB is greater than CD."
< is the symbol for "is less than."
36. A Theorem is a statement of which the truth is to be
proved. In proving the following theorem we use only
axioms and definitions, but in later theorems we may use
theorems of which the truth has been already established.
37a. Theorem: All straight angles are equal.
3
D
r
Given: BAG is a straight angle and EDF is any other
straight angle.
To Prove: ABAC = ZEDF.
Proof: Place the angle BAG on the angle EDF so that the
vertex A will fall on the vertex D, and side AC will fall
along DF.
But BC and EF are straight lines (14).
Furthermore, part of BC (AC) coincides with part of
EF (DF).
/. BC and EF coincide (8c).
/. Z BAG = Z EDF (35b).
37)). Theorem: All right angles are equal (33b).
38. Theorem: A straight angle contains 180°.
Proof: If a straight line is drawn through a point in a
plane it divides the sum of the angles about the point in the
plane into two equal parts — for the straight angle on the
one side of the point equals the straight angle on the other.
But the sum of the A about a point in a pl^ne = 360° (25) .
/. half of it = 180°. That is,
a straight angle = 180°.
Another way of stating the above theorem is :
39a. The sum of the angles about a point in a plane on one
16 GENERAL MATHEMATICS
side of a straight line (in the plane) passing through the poird
is equal to 180°.
39b. A right angle contains 90° (15, 38).
40. Theorem: If an angle contains 180 degrees , its sides
form a straight line.
Proof : For if it were possible to have a straight angle the
sides of which did not form a straight line, on comparing
this with a straight angle the sides of which did form a
straight line we should have two straight angles which were
not equal (35a).
But this is impossible (37a).
41a. The supplement of an angle is that angle which must
be added to it to make the sum equal to 180 degrees. That is,
'41b. Rule : To find the supplement of a given angle, sub
tract the given angle from 180 degrees.
42. The supplements of the same angle or of equal angles
are equal (32b).
43a. Supplementary Angles are angles whose sum is
equal to 180°, or a straight angle.
43b. Theorem: If two adjacent angles have their exterior
sides in a straight line, these angles are supplementary.
Given: The exterior sides AO and OB of the adjacent
angles AOC and COB are in the straight line AB.
To Prove: A AOC and COB are supplementary.
Proof: AOB is a straight line, (given)
/. I AOB is a st. Z (14).
GENERAL MATHEMATICS 17
But Z AOC+ Z COB= the st. Z AOB (32a).
/. the A AOC and COB axe supplementary (43a) .
Note: Compare this theorem with art. 39a.
43c. SupplementaryAdjacent Angles are adjacent angles
that are supplements of each other.
Exercise
Find the supplements of:
(1) (2) (3) (4) (5)
120°, 30°, 110°, 45°, 90°,
(6)
40° 50'.
180° may be written 179° 60'. The work may then be
arranged as follows:
179° 60'
40° 50'
139° 10' Ans.
(7) (8) (9) (10) (11) (12)
17° 21', 39° 42', 111° 27', 82° 37', 125° 57', 0°28',
(13)
47° 28' 13".
180° may be written 179° 59' 60". The work may then
be arranged as follows:
179° 59' 60"
47° 28' 13"
132° 31' 47" Ans.
(14) (15) (16) (17) (18)
2° 31' 5", 170° 19' 53", 158° 0' 2", 122° 59' 17", 1 6 0'7".
18 GENERAL MATHEMATICS
44. If the sides of an angle are prolonged so that they
form another angle having the same vertex, these two
angles are called Vertical Angles. Two intersecting straight
angles form two pairs of vertical angles.
45. Theorem: If one straight line intersects another straight
line, the vertical angles thus formed are equal.
Given: The lines M N and OP intersect at V.
To Prove: Z MVO = Z PVN.
Proof: Z OVM + Z M VP = 180° (43b).
Z PVN + Z MVP = 180° (43b).
That is: Both Z MVO and Z PVN are supplements of
Z MVP (41a).
/. Z MVO = Z PFJV (42). Q. #. 2).
By the same kind of proof show that Z M VP= Z OVN.
The preceding theorem (that is, the statement above the
figure) may be readily divided into two parts: what is
given and what is to be proved. This can be done with all
theorems: for example, in the first theorem (37a) it is given
that the angles are straight angles : to prove that they are
equal.
Notice that after a theorem is stated, a figure is drawn
and lettered to represent what is given in the theorem;
below this figure (after the word "given") must be definitely
stated what is true of the relation of the parts of the figure.
Nothing can be used in the proof which is not stated as
"given." For example, in the above theorem, if it were
stated as given, merely, that M N and OP were intersecting
straight lines — that is, if it were not stated that the lines
intersected at V, then no use could be made in the proof
GENERAL MATHEMATICS 19
of the fact that V is the vertex of the angles formed — for
that would imply that V was the intersection of the lines.
It is not necessary to state that MN and OP are straight
lines, for when a line is mentioned in mathematics it means
a straight line unless otherwise stated.
What is given is sometimes called the Hypothesis.
The student should always make sure that the last state
ment in the proof is that which we are directed to obtain in
the statement of which we are "to prove." For example,
in the above theorem we are to prove
Z MVO = Z PVN; and the last statement in this proof is
.\ Z MVO = Z PVN.
In other words, we have arrived at our answer.
To call attention to the fact that we have arrived at the
answer, it is customary to mark the last statement of a com
pleted proof with the initial letters of three Latin words
meaning "which was to be proved," that is, with the letters
Q. E. D., the initial letters of "Quod erat demonstrandum."
46a. Theorem: Two straight lines drawn from the same
point in a perpendicular to a given line, cutting off on the
given line equal lengths from the foot of the perpendicular, are
equal, and make equal angles with the perpendicular.
Given: AF J_ to GL at F, P any point on AF, CP and
PD two straight lines drawn from P, cutting off on GL the
equal lengths CF and FD.
To Prove: CP = PD; Z CPF = Z FPD.
Proof: Fold the left hand half of the figure over, along
20 GENERAL MATHEMATICS
the line AF, until it falls on the plane at the right of AF, and
its plane coincides with that of the right hand half (7b).
Z AFG= Z AFL (37b)
Then, since side AF coincides with side AF and the vertex
F with the vertex F, side GF will fall along FL (35a).
C will fall on D (CF= FD, given).
Then, since P is common to both CP and PD, CP coin
cides with PD (8c).
Also, PD = CP (8d).
Again, since PF coincides with PF, P with P and PC
with PD, Z CPF = Z FPZ) (35b). Q. E. D.
46b. A Segment is a portion of a line included between
two definite points. (Thus, in the above figure, CF is a
segment of GL.)
47a. To Bisect means to divide into two equal parts.
47b. The Perpendicular Bisector of a given line is that
perpendicular to it which divides it into two equal parts.
48a. A Postulate is a construction admitted to be possible.
The following are postulates.
48b. A straight line can be drawn between any two points
(8b).
48c. A straight line can be produced indefinitely (8g).
48d. Any point in a plane may be taken as a center of a
circle of any radius.
For example, by the first postulate (48b), if A and G were
given points in the preceding figure, we might draw the
line AG but it would not be in accordance with the postulate
to attempt to draw a line from A to G equal to AF — the
postulate does not state that a line of a given length can be
drawn between any two points.
Let the student beware of attempting to make a construction line fulfill
two relations at once (as in the case cited above) ; if two are necessary it
should be assumed that the line fulfills only one of them, and then the other
must be proved.
GENERAL MATHEMATICS 21
49. Theorem: All points in the perpendicular bisector of a
given line are equally distant from, the extremities of the line,
and no point outside of the perpendicular bisector is equally
distant from the extremities of the line.
Given: GL is a straight line, BI is the perpendicular
bisector of GL at / the middle point of GL; P any point
in BI, X a point outside of BI.
B
X
•
c,
s
< \
//
*?N \
/ y
X\ \ .
/.••
\\ \
//
X:,\
>y
V.
s
%•
.1*
•'*
.»
\
To Prove: (1) GP= PL;
(2) GX is not equal to XL.
Proof: (1) Draw PG and PL.
GI = IL (given).
.. GP = PL (46a).
(2) Draw XG and XL.
Since X is not in BI, one of these lines must intersect BI.
Let XG intersect BI at C.
Draw CL
XC+CL> XL(8f).
But CL = CG (Part 1).
Substituting for CL its equal in the above inequality:
XC+CG> XL
That is, • GX > XL (32a).
In other words, since X, any point outside of BI is un
equally distant from G and from L, all points outside of
BI are unequally distant from G and from L, that is, no
point outside of BI is equally distant from G and from L.
22 GENERAL MATHEMATICS
60. A line added to a figure to assist in the proof is called
a Construction Line. The student should be careful that
every construction used is a postulate. A construction line
should always be dotted, as G X in the preceding figure.
51. To Determine a line means to locate it definitely.
52. Theorem : Two points equidistant from the extremities
of a straight line determine the perpendicular bisector of the line.
Proof : These two points cannot lie outside of the perpen
dicular bisector (49)
/. they must he in the perpendicular bisector.
A line joining these two points must coincide with the
perpendicular bisector (8c). Q. E. D.
To construct the perpendicular bisector of a given line.
aJ.h
w : c
•
•
•
•
•
•
•
•
Given: GL is a straight line.
To Construct: the perpendicular bisector of GL.
Construction: With GP, any length greater than half GL,
as a radius, and G as a center, construct arcs AC and DE.
(I) Any point in AC and DE will be at a distance GP
fromG (18).
With GP as a radius and L as center construct arcs FH
and JL, intersecting AC and DE at P and B, respectively.
(II) All points in FH and JL will be at a distance GP
from L. Why?
But P lies in both AC and FH.
:. P is at a distance GP from both G and L (I, II).
That is, P is equidistant from G and L.
GENERAL MATHEMATICS
23
In like manner, B is equidistant from G and L.
.'. PB is the perpendicular bisector of GL (52).
The student may now understand the following:
53. Rule : To construct the perpendicular bisector of a given
line; with one extremity of the given line as a center and any
radius greater than half the line describe arcs of the same circle
both above and below the line; with the other extremity as a
center and the same radius describe arcs intersecting the first,
and draw a straight line between the points of intersection. This
straight line will be the perpendicular bisector of the given line.
Note: This construction, and those which follow, are to be used by the
student in the blackboard construction of figures. Make no figures freehand.
To erect a perpendicular to a given line at a given point in
the line.
•H
Ai
JF
•ft
Given : P any point in the straight line GL.
To Construct: MN perpendicular to GL at P.
Construction: With any radius as 4P and with P as a
center describe arcs intersecting GL at A and B.
Construct MN the perpendicular bisector of AB (53).
P is equidistant from A and B (16).
/. Plies in MN (49).
That is, MN is ± to AB at P.
That is, MN is J_ to GL at P (12a, 10).
The student may now understand the following:
24
GENERAL MATHEMATICS
54. Rule: To erect a perpendicular to a given line at a
given point in the line; with the given point as a center and any
radius describe arcs intersecting the given line on either side
of the given point; with these points of intersection as centers
and a longer radius describe arcs intersecting above and below
the line; a line connecting these points of intersection will be the
perpendicular desired.
55. Theorem: At a given point in a given straight line,
there can be but one perpendicular to the line.
Given: F any point in the straight line GL and PF 1
to GL.
To Prove : PF is the only perpendicular to GL at F.
Proof: If possible let XF be a perpendicular to GL at
F.
Then Z PFL is a rt. Z and Z XFL is a rt. Z.
/. Z PFL = Z XFL (37b)
But this is impossible (33a) .
.\ PF is the only ± to GL at F. Q. E. D.
From a given external point to let fall a perpendicular upon
a given line.
\
;r
*
*
aV
•
\ i /
J**
' c %
.b
GENERAL MATHEMATICS 25
Given: GL is a straight line, P any external point.
To Construct: A J. to GL from P.
Construction: From P as a center, with any radius,
draw arc MN, intersecting the given line at A and B.
P is equidistant from A and from B (18).
With A as a center and any distance greater than one
half of A B as a radius describe an arc, and with £ as a
center and the same radius describe an arc intersecting the
arc just drawn at C. Draw PC.
PC is the perpendicular bisector of AB (52).
That is, it is perpendicular to GL (12a, 10).
The student may now understand the following:
56. Rule : From a given external point to let fall a perpen
dicular upon a given line: with the given point as a center and a
radius of sufficient length describe an arc intersecting the given
line in two points; with these points of intersection as centers,
and a radius greater than half the distance between them,
describe intersecting arcs. A straight line through the point
of intersection of these arcs and through the given point will be
the de&ired perpendicular.
57. Theorem : Only one perpendicular can be drawn from
a given external point to a given straight line.
Given: GL any straight line, P any external point, PF
a perpendicular from P to GL, PA any other line drawn
from P to GL.
To Prove: PA is not J_ to GL.
26 GENERAL MATHEMATICS
Proof: Produce PF to E, making FE = FP. Draw AE.
PE is a straight line. (Construction)
.% PAE is not a straight line (8a).
:. Z PA J? is not a st. Z (14).
Since PE is J_ to GL, GL is i. to PE (12b) and by con
struction PF = P#.
/. Z PA* 7 = Z FAE (46a).
That is, Z PA* 7 = 1/2 Z PA# (32a).
But PAE is not a st. Z .
.'. half of it is not a rt. Z .
That is, Z PAF is not a rt. Z .
.\ PA is not J_ to GL.
But PA is any line other than PF drawn from P to GL.
:. PF is the only ±. Q.E. D.
The student is now ready to consider the question ""What
is a straight line?" Although the student knows well
enough, practically, what a straight line is, so that he would
not mistake a straight line for a curved line, he will yet
find it hard to compose a good mathematical definition of a
straight line. A statement of an axiom about a straight
line is not a definition of it, any more than the statement that
"two points equidistant from the extremities of a straight
line determine its perpendicular bisector" is a definition of
a perpendicular bisector. It is therefore incorrect to state
"a straight line is the shortest distance between two points"
as a definition of a straight line.
To understand the following definition of a straight line
let the student consider the accompanying figure.
 * J* c g S L
If from anywhere in the straight line GL a segment is cut,
as AB y this segment may be placed anywhere on GL, as
GENERAL MATHEMATICS 27
at CD, and made to coincide with the line completely, pro
vided both extremities, A and B, lie in the line. The
segment AB would not, of course, coincide with GL if placed
in the position ED or HF. Furthermore, not only will AB
coincide with the line as first placed, in the position CD
(with A at C and B at D), but it also will coincide with
the line if reversed, so that A falls at D, and B at C, that is
AB can be made to coincide with GL, however placed on
GL, provided A and B lie in GL.
The student may now understand the following definition :
68. A straight line is a fine of which any segment, however
placed on any other part of the line, will coincide with that
part if its extremities lie in that part.
The theorems in this and following chapters are called
Theorems in Plane Geometry. Plane Geometry is the study
of Plane figures.
II. Field Exercise
LAYING OUT FOUNDATION OF BUILDING
Equipment: Level, ranging pole, knotted cord,* four
stakes (3 ft. long), a dozen stakes (18 in. long), mallet, ham
mer, nails.
Personnel: Leader, one student to use level, one to
hold ranging pole, one to drive stakes; three to hold knotted
cord (one at each end and one at middle).
Procedure : Drive a long stake A to mark front corner of
building. Place a stake B twenty feet to one side and here
set up level so that observer may look through instrument,
across corner, and along proposed front of building. The
ranging pole should be held a short distance from A (on the
*The knotted cord is prepared by tying a knot at the center of a piece of
cord about 100 feet long, then doubling the cord and cutting off the two ends
so that they are exactly even; that is, so that the knot comes in the middle.
At each end of the «ord make a loop to serve as a handle. (These loops should
be of the same size.)
28
GENERAL MATHEMATICS
side opposite the level) and directed by the observer until it
is exactly in the line BA produced (Field Exercise I, Use
of Level).
This position should be marked with a stake. Continue
this process until a line has been staked out at least twenty
feet longer than the desired front of the building.
A
;
I
f
f
•
f
t
•
§
t
3
%
t
%
Starting from A measure along this line of stakes the dis
tance desired for the front of the building, and mark the
end of this distance with the long stake D. A and D are
the two front corners of the building.
To locate the two rear corners, proceed as follows: First
erect a perpendicular to the line AD at A, in this manner;
measure the distance BA, and locate another stake E the
same distance to the other side of A — marking this stake
(with pencil, or with a sheet of paper through which the top
of the stake is thrust) to avoid confusing it with the other
stakes. By construction, A is equally distant from B and
E. One end of the knotted cord should be held at E y the
other at B, and the cord drawn tightly by the knot in the
middle. This knot should be marked by the stake F. FA
GENERAL MATHEMATICS 29
is perpendicular to AD (54). (If it is desired to prolong
AF, this may be done by setting up the instrument at A,
and sighting across F.) From A measure toward F the
distance desired for the side of the building, and mark it
with a stake G.
By the same method a stake H should be located for the
other rear corner. To test the work the distance GH
should be measured; if the length GH is the same as AD,
the work is correct. To assist in measuring these distances,
a nail should be driven in the center of each stake.
All the short stakes may now be taken up, and the four
corner stakes, A, D, H , and G, driven in firmly.
Note: If the Farm Level is not completed, the Water Level may be used
instead. Since the instrument is here used, not to lay off levels, but simply
straight lines, it is not necessary to float it in a pail of water, but simply place
it on top of a box or other support. Instead of the water level, a three foot
length of water pipe (about an inch in outside diameter) may be used. The
pipe should be supported between two stakes, with Vshaped notches in their
tops, driven into the ground two feet or more apart.
CHAPTER III
PARALLEL LINES
69. Parallel Lines are lines in the same plane which cannot
meet, however far they may be produced.
The symbol for parallel is //.
The symbol for parallel lines is //s.
Parallel lines are often called "Parallels."
60a. Theorem: Two or more straight lines in the same
plane, perpendicular to the same straight line are parallel.
6
Given: FP and AB ± to GL.
To Prove: FP and AB are//.
Proof: If FP and AB were not //, they would meet at
some point. Then, from this point there would be two lines,
FP and AB, both ± to GL.
But this is impossible (57).
.\ FP and AB do not meet — that is, they are/// Q. E. D.
Compare the above method of proof with that used in
Art. 55.
Note: It is upon this principle that parallel lines are
drawn with the Tsquare — by sliding the crosspiece of the
Tsquare up and down the edge of the drawing board.
30
GENERAL MATHEMATICS 31
60b. To construct a parallel to a given line through a given
external point.
C
r
Given: AB a, straight line and P a given external point.
To Construct: A line through P, // to AB.
Construction: From P let fall a J_ PF upon AB (54), and
produce this line upward through P.
At P erect a J_ (CD) to PF (54).
CD will be// to AB (60a).
In the same manner any number of perpendiculars to PF
may be proved parallel to each other.
6ia. Axiom: Through a given point only one straight line
can be drawn parallel to a given straight line.
61b. Two or more straight lines in the same plane, parallel
to the same straight line are parallel to each other.
A B
C D
G L
Given: AB and CD // to GL.
To Prove : AB // to CD.
Proof : If AB and CD are not // they will meet at some
point. Then from this point there would be two lines //
to GL. But this is impossible (61a) .
.\ AB and CD cannot meet and so are //. In like man
ner, the theorem could be proved to hold for any number
of //s. Q. E. D.
Let the student prove this theorem to hold for three lines
// to a given straight line.
32
GENERAL MATHEMATICS
Note: In the following proof it may at first seem absurd to draw the Use
MN and consider it as being perpendicular to PR. It should be borne in
mind, however, that the line MN symbolizes a relation about which we are
reasoning. To aid us in the proof, we are to consider a line through C per
pendicular to PR and it is necessary to keep this idea distinct from the idea
of a parallel to GL drawn through C.
62. Theorem : // a straight line is perpendicular to one oj
two parallel lines it is perpendicular to the other also.
n*
TV
Given: GL and AB are two parallel straight lines, PR a
perpendicular to GL at F, cutting AB at C.
To Prove: PR is ± to AB also.
Proof: Through C draw MN ± to PR.
(Since we do not yet know whether AB is perpendicular
to PR or not, we do not draw MN coinciding with AB.)
M N is // to GL (60a).
/. MN coincides with AB (61a).
But M N is ± to PR (Construction).
/• AB, which coincides with AfiV, is also ± to PR.
That is, PR is ± to AB. Q. E. D.
63a. A Transversal is a straight line which cuts across
two or more other lines.
63b. The angles formed on the inside of two given straight
lines cut by a transversal on opposite sides of the transversal,
one adjacent to one of the given straight lines and the other
to the other given straight line, are called Alternate Interior
Angles.
GENERAL MATHEMATICS
33
The abbreviation for alternate interior angles is "Alt.
Int. A."
TN is the Transversal, AB and CD are the given straight
lines. '
Thus, Z BEF and Z EFC are Alt. Int. A: Z AEF and
Z EFD are Alt. Int. A
Note: "Alternate/' the first part of the name, indicates
that the angles are opposite sides of the transversal, while
"Interior" indicates that they are inside the given straight
lines.
64. Theorem: If two parallel lines are cut by a transversal
the alternate interior angles are equal.
Given : GL and PR are two parallel lines cut by the trans
versal TN at points A and B.
To Prove: Z PBA = Z BAL.
Proof: Through C, the middle point of AB } draw DE ±
to PR. Then DE is ± to GL also (62).
(I) That is, DB is J. to DE.
34 GENERAL MATHEMATICS
Also the fact that DE is ± to AE may be stated as AE
is ± to DE (12b).
Fold the figure directly upwards about the point C, so
that CE falls along CE produced, that is, along CD (8g).
Then AE, which is ± to CE, is ± to CD. (CE now coin
cides with CD.)
Now fold the figure ACE to the right about CE (which
coincides with CD) as an axis until A falls upon the plane
to the right of DC.
Since CE still coincides with CD,
(II) AE (which is ± to CE) is J_ to CD.
Also, since Z ACE = Z DCB (45), and C falls on.C, AC
will fall along CB (35a).
But AC = CB. (C is middle point of AB.)
:. A falls on B.
. Now both AE and DB are ± to CD (I, II).
.". DB and AE coincide (57), and since A coincides with
B, and AC coincides with CB,
. Z PBA = Z SAL (35b). Q. E. D.
66. Theorem: // the alternate interior angles, formed by a
transversal intercepting two straight lines are equal, the two
straight lines are parallel.
5—
Given: GM and SL two straight lines, cut by the trans
versal TN at points A and B, so that Z GBA equals Z
BAL.
To Prove : SL is // to GM.
GENERAL MATHEMATICS 35
Proof: Consider a line drawn through A // to GM .
Since we do not yet know whether this line coincides with
SL or not, we will consider it, temporarily, as occupying the
position CD.
Z GBA = Z BAD (64).
But Z GBA = Z BAL (Given).
.. Z BAD = Z BAL (31),
Since ^ BAD and BAL have side BA in common, and
vertex A in common, AZ> and AL coincide (35a).
.". SL and CD coincide (8c) .
That is SL coincides with a line // to GM and so SL
is // to GM. Q. E. D.
Class Room Exercise
Take a piece of paper with parallel sides (such as an
ordinary small blotter) and cut it across slantingly with the
scissors. The angles thus formed may be made to coincide
by reversing one part of the blotter and placing it over the
other. What theorem does this illustrate?
66. When two given straight lines are cut by a trans
versal, a pair of angles on the same side of the transversal,
but with one of the angles inside the given lines and the
other outside the lines, are called ExteriorInterior Angles.
Thus in the preceding figure, NBG and BAS are exterior
interior angles.
The name "ExteriorInterior Angles' ' means "outside
inside angles" since one angle X>i the pair is outside the given
lines and the other inside.
Thus when a transversal intersects two parallels, it forms
but two pairs of alternateinterior angles, but forms four
pairs of exteriorinterior angles. Thus, in the preceding
figure, the pairs of exteriorinterior angles are as follows:
NBG and BAS f GBA and SAT, NBM and BAL, MBA
and TAL.
36
GENERAL MATHEMATICS
67a. Theorem : If two parallel lines are cut by a transver
sal, the exteriorinterior angles are equal.
Given: PA and R L are two parallel lines cut by the
transversal TN at B and C.
To Prove: Z TBA = Z BCL.
Proof: Z TBA = Z PBC (45).
Z PBC = Z BCL (64).
.\ Z TBA = Z BCL (31). Q. E. D.
In like manner it may be proved that Z TBP = z BCR.
Also Z PBC = Z RCN, etc.
Name the other pair of angles in the above figure which
are equal by this theorem.
67b. If the exteriorinterior angles formed by a transversal
intersecting two straight lines are equal, the two straight lines
are parallel.
GENERAL MATHEMATICS
37
Given: AB and CD two straight lines cut by the trans
versal TN at points E and F so that Z TEB = Z EFD.
To Prove : AB is// to CD.
Proof: Z A.EF = Z 7\E.B (45).
Z TEB = Z .Etf'D (Given).
.♦. Z A^F = Z EFD (31).
.. A.B is // to CD (65). Q. E. D.
#
Ilia. Shop Exercise
The Parallel Ruler is a useful instrument for drawing lines
parallel to a given line. As seen from the diagram, the
distance between the bars can be varied at will.
rt.W.f
r
The parts represented are :
A A y strips of wood 1/4 in. X 1 in. X 12 in.
BB, strips of brass, 4 in. long, with holes for rivets.
After constructing this instrument the student can easily
design and execute a parallel ruler of larger size (using
wooden strips instead of brass), for use at the blackboard.
Illb. Field Exergise
MARKING CORNERS BY BATTER BOARDS
Since it would not be possible to excavate for the founda
tion thus laid out (II) without disturbing the stakes mark
ing the corners, it is customary to transfer these corner
marks to "Batter Boards." As shown in the diagram (Illb)
a Batter Board is formed of two pieces of board nailed to
38
GENERAL MATHEMATICS
three or four pieces of upright studding so as to form a right
angle. The upper edges of these boards should be planed
smooth. A Batter Board should be placed at each corner
of the foundation to be marked.
Equipment: Water level (with pail of water in which
to float level), box (upon which to Test pail) ; ball of fishline
or other strong cord; 8 pieces of board, 3 feet long, planed
on one edge; 12 pieces of studding, 2 inches wide, 3 inches
thick, and 5 feet long; half pound sixpenny nails; hammer;
plumbbob.
Procedure: About two feet outside each corner stake,
drive in the uprights, as shown in the diagram. The first
GanHter* &oardi
3trtm
board should be fastened to two of these uprights at
one end of the building, at a front corner. At first only
one end of the board should be fastened to an upright
(and that with a single nail) at the height desired for the
masonry foundation. Locate the water level (in the pail
of water, on top of the box) so that the line of sight crosses
this board and traverses the front of the building. (The line
of sight may be raised by pouring more water into the pail.)
While the student at the water level moves it slightly from
GENERAL MATHEMATICS 39
one side to the other, so that his line of sight sweeps along
the top of the board, another student should .move the un
nailed end of the board up or down until it is level, and
then nail it. Then (under the direction of the student at
the water level), the board parallel to this, at the other front
corner of the building, should be nailed at exactly the same
level as the first board.
The water level should now be moved so that the board
forming a right angle with that first located may be adjusted.
(To get the fine of sight in the same level as before, it is best
to pour out some of the water, and then add water gradually .
until the line of sight reaches the same level shown by the
board first nailed.) By moving the water level from one
side to the other, nail this board exactly level with the board
first placed, and forming a right angle with it. A board
parallel to this, and at the same level, should now be located
at the rear corner of the building.
The water level should now be moved to the other front
corner of the building and the process repeated.
The batter boards on the two front corners are now com
plete — those at the rear corners still lack the end boards.
These should now be put in position, each of these end
boards being made level with its adjacent side boards. The
work may now be tested by sighting along the length of the
building with the water level; if the work has been accu
rately done, these two boards last located should be at the
same level.
A student should now station himself at each of the front
batter boards — holding one end of a length of cord so that
it is parallel to the front of the building, and drawing it
tight over the end boards. Another student, with the
plumbbob, should direct them in moving the cord so that
it is directly above the nails in the corner stakes (II).
* Then nails should be driven carefully in the sides of the end
boards (not the upper edge, as the hammering tends to dis
40 GENERAL MATHEMATICS
turb the level), and the cords fastened to them, so that they
are stretched* taut. In the same way, a cord should be
fastened at the rear of the building and also at the two ends.
As shown in the diagram, the intersections of these cords
mark the corners of the foundation, while the planed edges
of the board mark the level for the masonry construction.
The corner posts (II) may now be removed, leaving the
ground clear for excavation. .
Note : Instead of the water level, the water pipe substi
tute (II) may be used, in combination with a carpenter's
level — the pipe being brought to the proper alignment by
driving in the stakes. In this case, for leveling the second
board of the batter boards, the carpenter's level is applied
directly to the upper edge of the board.
CHAPTER IV
TRIANGLES
68a. A Triangle is a portion of a plane bounded by three
straight lines.
68b. The lines bounding a triangle are called its sides,
and the points of intersection of the sides are called the
vertices of the triangle. The symbol for triangle is A.
69. Any side of a triangle may be called its base; the
perpendicular let fall from the opposite vertex to the base
is called the altitude of the triangle.
70. An Isosceles Triangle is a triangle having two equal
sides.
71. Two triangles are equal if the three vertices of the one
coincide with the three vertices of the other, that is, two tri
angles are equal if they may be made to coincide throughout (8c).
72a. The corresponding angles of two equal triangles are
angles opposite equal sides.
72b. The corresponding sides of two equal triangles are
sides opposite equal angles.
72c. Corresponding parts of equal triangles are equal.
Note : Corresponding parts are sometimes called Homol
ogous parts.
73. To construct a triangle equal to a given triangle.
42 GENERAL MATHEMATICS
From any vertex of the given triangle (as C) let fall a per
pendicular to the opposite side.
This perpendicular CD will then be the altitude of the trian
gle and the side AB to which it is drawn will be the base (69):
to construct an equal triangle lay off a straight line equal to
the base; at a point which is the same distance EH from
one extremity E of this line as the foot of the altitude of the
given triangle is from the extremity A of its base, erect a
perpendicular (54); lay off on this perpendicular GH a dis
tance equal to the altitude of the triangle; lines joining point
G to the extremities of EF form with EF the triangle desired.
The proof of this is left to the student.
74. Theorem : The sum of two sides of a triangle is greater
than the third side.
Given: In the triangle CAB let AB be the longest side.
To Prove: AC + CB > AB.
Proof: AC + CB > AB (8f). Q. E. D.
Note: Since AB is the longest line, there is no need of
going through a proof to show that AB + BC > AC, or that
AB + AC > BC.
Lines which are equal may be marked with the same
number of cross lines, and angles which are equal by an
equal number of small arcs. Thus, in the next figure, the
angles B and E, being equal, are marked with one arc, and
C and F with two arcs. In the succeeding figure AB and
DE, being equal, are marked with one cross line, and EF
and BC, being equal, are each marked with two cross lines.
GENERAL MATHEMATICS
43
76. Theorem: Two triangles are equal if two angles and the
included side of the one are equal to two angles and the included
»ide of the other.
Given: In A ABC and DEF, BC = EF, Z B = Z E,
and Z C = Z F.
To Prove: A ABC = A DEF.
Proof: Place Z DEF on Z ABC so that EF coincides
with its equal BC.
DEwW. fall along AB ( Z E = Z B, given).
D/? will fall along AC ( Z F = A C, given).
Now, AB and AC intersect at A, therefore DE and DF
which coincide with them must intersect in the same point.
.. D will fall on A (8e).
/. A DEF = A ABC (71). Q. E. D.
76. Theorem: Two triangles are equal if two sides and the
included angle of the one are equal to two sides and the included
angle of the other.
Given: In A ABC and DEF, DE = AB, EF = BC
and Z E = Z B.
To Prove: A DEF = A ABC.
44
GENERAL MATHEMATICS
Proof: Place A DEF on A ABC so that EF coincide?
with its equal, BC. Then, since E falls on B, ED will fal
along AB ( Z E = Z B, given).
D falls on A (ED = AB, given).
Then, since F falls on C, DjP coincides with AC and if
equal to it (8c, 8d).
.\ A DEF = A ABC (71). Q. E. D.
77a. To say that two lines are equal by identity, means
that they are really the same line.
77b. Theorem : In an isosceles triangle the angles opposiU
the equal sides are equal.
Given: Let ISC be an isosceles triangle, having IS
SC.
To Prove: Z I = Z C.
Proof : Let SF be drawn 90 as to bisect the Z S.
In A SIF and SFC f
SI = SC
SF = SF
Z ISF= Z FSC
A SIF = A SFC
/. z I = z C
(Given)
' (Identity)
(Construction)
(76)
(72c) Q. E. D.
GENERAL MATHEMATICS 45
78a. Theorem: Two triangles are equal if the three
sides of the one are equal to the three sides of the other.
Given: In the A ABC and DEF, let DE = AB, DF =
AC, and EF = BC.
To Prove: A DEF = A ABC.
Proof: Place A DEF below A ABC, as shown by the
dotted lines, so that EF coincides with its equal BC, D fall
ing on the opposite side of BC from A, E coinciding with B
and F with C. Draw AD.
BA = BD (given).
/. A ABD is isosceles (70).
I. .. Z BAD = Z BDA (77b).
Again, AC = DC (given).
.\ A ACD is isosceles (70).
II. /. Z CAD = Z CD A (77b).
Adding equations (/ and II)
Z BAD + Z CAD = Z J5DA + Z CDA (34b).
/. Z J5AC= Z J5DC(32a).
That is, ZBAC= AEDF.
.. A AJ5C = A DEF (76). Q. E. D.
To bisect a given angle.
A
Given: AJSC any angle.
46 GENERAL MATHEMATICS
To Construct: PB bisecting ABC.
Construction: With B as a center, and any convenient
radius, describe an arc intersecting AB and BC at E and F,
respectively.
With E as a center and any convenient radius (greater than
half the distance EF), draw an arc, and with F as a center,
and the sajne radius, draw an arc intersecting the arc just
drawn at G. BG (produced as much as desired) is the bisector
of Z EBF.
Proof: EB = BF (18).
EG = GF (18).
GB = GB (Identity).
.\ A GBE = A GBF (78a).
.\ Z EBG = Z GBF (72a).
The student may now understand the following:
78b. Rule: To bisect a given angle: with the vertex of the
given angle as a center, and any radius, draw an arc intersecting
the sides of the given angle: with these points of intersection as
centers, and any length greater than half the distance between
them as a radius, draw intersecting arcs; a line drawn from
this point of intersection to the vertex will be the bisector of the
angle.
Blackboard Exercise
By aid of Art. 53 (or Art. 56) and 78b, construct an angle
of 45°.
79a. To suggest that points or lines have some important
relation, they may be designated by the same letters — these
letters of the second point being distinguished from those of
the first point, or line, by a small mark placed above and
to the right of the letter, which is read "Prime." Thus in
the following figure, the segment E f F' of GL is so marked
to indicate that it is equal to EF.
E 1 F r is read, "E prime F prime."
GENERAL MATHEMATICS 47
79b. To construct a triangle when three sides are given.
0 c DEr
£ m • r  •
Let the given lines be AB, CD, and EF.
Draw a line of indefinite length, as GL; on GL lay off a seg
ment equal to one of the given lines — as E f F f equal to EF.
With one of the remaining lines as a radius, as AB } and E r
as a center, draw an arc, as HI.
With the remaining, given side as a radius, and f'asa cen
ter, describe an arc J K, intersecting the first at P.
Then, since P is in the arc HI it is at a distance AB
from E' and since it is in the arc JK it is at a distance CD
from F' (18).
Draw PE' and PF'.
The A PE' F' is the A desired.
For PE'= AB, PF'=CD and E'F' = EF.
The student may now understand the following:
79c. Rule: To construct a triangle when the three sides of
the triangle are given: lay off a length equal to one of the given
sides — then with another of these sides as a radius and an ex
tremity of the line just laid off as a center , draw an arc; then
with the third side as a radius and the other extremity of the
line first laid off as a center, draw an arc intersecting the arc
first drawn; lines drawn from the point of intersection of these
arcs to the extremities of the line first laid off will form with this
line the triangle desired.
48 GENERAL MATHEMATICS
80a. Theorem : The sum of the three angles of a triangle is
equal to 180 degrees.
Given: ABC is a triangle.
To Prove: AA+ AB + ZC= 180°.
Proof: Through B draw PL // to AC.
(I) A PBA + A ABC + ACBL = 180° (39a)
APBA = ABAC (64)
ALBC = ABC A. (64)
Substituting these values in (I)
ABAC + A ABC + ABC A = 180°. Q. E. D.
80b. This theorem may also be stated as follows:
The sum of the three angles of a triangle is 2 rt. angles.
80c. // two angles of one triangle are equal to two angles
of another, then the third angle of the first triangle equals the
third angle of the second.
Proof: For the third angle of the first triangle is the sup
plement of the sum of the two given angles of that triangle
(80a; 41b) and the third angle of the second triangle is the
supplement of the sum of the two given angles in that tri
angle.
/. The third angle of the first triangle equals the third
angle of the second triangle (42). Q. E. D.
The above theorem (80a) was known long before the
Christian era. Aristotle, who lived about 350 B. C, referred
to it frequently.
Let the student prove the above theorem by drawing the
construction line through C parallel to AB.
GENERAL MATHEMATICS
49
80d. Theorem: The perpendicular from the vertex to the
base of an isosceles triangle, bisects the base and the angle at
the vertex.
Given: AB and AC are the legs of the isosceles triangle
ABC; AF the ± from the vertex A to the base BC.
To Prove: BF = FC; Z BAF = Z FAC.
Proof: In A BAF and FAC
Z B = Z C 
Z BFA = Z CFA
Z BAF = Z FAC
But BA = AC
AF = AF
A BAF = A FAC
:. BF = FC ■
(77b)
(37b)
(80c)
(Given) .
(Identity)
(76)
(72c) Q. E. D.
81a. If a side of a triangle is produced, the angle thus
formed outside of the triangle, between the side produced
and a side of the triangle, is called the Exterior Angle of the
triangle.
Exercise
Let the student prove the following:
The bisectors of the equal angles of an isosceles triangle,
together with the base, form an isosceles triangle.
50 GENERAL MATHEMATICS
81b. Theorem: An exterior angle of a triangle is equal to
the sum ol the two opposite interior angles.
Given: A BCD is the ext. Z of A ABC.
To Prove: Zi + ZB= ABCD.
Proof: The sum of A A + A B is the supplement of angle
BCA (41b).
[For AA+ AB + A BCA =180° (80a)].
Now, ABCD is supplement of A BCA (43b).
.. Z BCD = the sum of Z A + Z B (42). Q. E. D.
Since an exterior angle is equal to the sum of the two op
posite interior angles it follows that:
81c. An exterior angle of a triangle is greater than either
of the two opposite interior angles.
IVa. Class Room Exercise
With a pair of scissors cut out a triangle of paper; tear
off the corners and lay them on the desk so that the straight
edges just touch each other. The outer edges will form a
straight line (80a).
Repeat the exercise with triangles of different shapes.
GENERAL MATHEMATICS 51
IVb. Field Exercise
MEASURING THROUGH AN OBSTRUCTION
Equipment : Levelinginstrument, rangingpole, stakes, cord.
Procedure: Suppose it is desired to measure the dis
c
tance along the edge of a field from A to B and this line is
obstructed by a thick growth of trees (C).
Locate a stake D at some point in the line AB, near the
obstruction, and another stake E on the opposite side of the
obstruction, and in the line AB. Set up the leveling instru
ment at some convenient point as F, and stake off the straight
line FE (Field Exercise II) . Set up the instrument again at
some other point G and stake off the line DG, intersecting the
first line at a point marked with the stake H. (To aid in
locating this point exactly, hold the cord along the . stakes
in one line, and the tape along the stakes in the other.)
Measure DH and lay off the same distance from H to
ward the point (?, marking the end of this distance with the
stake /. Measure EH and lay off the equal length HJ.
Measure J I and this will give the length DE.
For DH = HI and ## = HJ (by construction).
Z DHE= Z JHI (45).
/. ADHE = A JHI (76).
/. JI = DE (72c).
CHAPTER V
CIRCLES
Review Arts. 16 to 27 inclusive.
O is the symbol for a circle. (§) is the symbol for circles.
82a. Two circles are equal if they have equal radii.
For they will coincide if their centers are made to coin
cide. Hence:
82b. Two equal circles have equal radii.
82c. Concentric Circles are circles having the same center.
83. The converse of a theorem is obtained by interchang
ing what is given with what is to be proved.
The two following theorems are converses of each other:
84a. Theorem : In the same circle or in equal circles , eqvd
central angles intercept equal arcs.
Given: In the equal ® BAM and CDP the centers of
which are and N, the central A BOA and CND are equal.
To Prove : Arc B A = arc CD.
52
GENERAL MATHEMATICS
53
Proof: Place the circle CDP on circle BAM so that the
Z.CND shall coincide with its equal ZBOA, that is, so
that N falls on 0, ND along OA and CN along BO.
Then D falls on A and C on £ (82b).
.\ Arc CD coincides with arc BA (16). Q. E. D.
Note: By Art. 16, no part of CD may fall inside of BA
or outside of BA.
84b. In tAe same circle or in equal circles, equal arcs are
intercepted by equal central angles.
Given: (in figure of 84a) Arc AB = arc CD.
To Prove: Z AOB = Z DNC.
Proof: Place O DCP on O ABM so that DN shall fall
on its equal AO and arc DC on its equal AB (16).
Then: since C falls on B, and iV on 0, NC will coincide
with OB (8c).
That is, the Z DNC will coincide with the Z AOB.
.. Z A0J5= Z D^C (35b). Q. #. Z).
85a. A chord of a circle is a straight line having its ex
tremities in the circumference.
85b. The arc intercepted by the straight line is said to
be subtended by it.
86a. Theorem : In the same circle or in equal circles , equal
arcs are subtended by equal chords.
Given: In the equal ® the centers of which are and
C, let arcs AB and ED be equal.
54
GENERAL MATHEMATICS
To Prove: Chord AB = chord DE.
Proof: Draw radii OA, OB, CD, and CE.
OA = CE (82fo)
OB = CD (82b)
Z 40B = Z ECD (84a)
.. A AOB = A DCE (76)
.*. Chord AB = chord D^ (72c).
86b. Theorem: Conversely, in the same circles or in tqd
circles, equal chords subtend equal arcs.
Given: In the equal (?) (Tig. of 86a) the centers of which
are and C, the chords AB and DE are equal.
To Prove: Arc AB = arc DE.
Proof: Draw radii OA, OB, CD, CE.
In A OAB and .DOE,
04 = CE
0B = CD
Chord 4B = chord DE
.'. A AOB = A CZ>£
.*. Z 40B = Z DOE
.*. Arc AB = arc ED
At a given point in a given straight line to construct an
angle equal to a given angle.
(82b)
(82b)
(given)
(78a)
(72c)
(84a). Q.E.D:
Given: CAB any given angle; P a point in the straight
line GL.
To Construct: An Z equal to Z CAB, having the vertex
P and the side PL.
GENERAL MATHEMATICS 55
Construction: With A as a center and AB as a radius
draw arc CB, and with P as a center and same radius draw
arc DE.
Arc CB and arc DE are arcs of equal circles (82a).
Measure chord CB and lay off same distance FE from E
on arc DE.
Arc FE = arc CB (86b).
/. Z FPE = ZCAB (84b).
The student may now understand the following:
87. Rule: At a given point in a given straight line to con
struct an angle equal to a given angle; with the vertex of the
given angle as a center and any radius draw an arc inter
secting both sides of the angle; with the given point as a center
and the same radius draw an arc intersecting the given line;
set the compasses to measure the chord of the arc intercepted
between the sides of the given angle and lay off an equal chord
on the arc just drawn, beginning at its intersection with the
given line. A line drawn from the other end of this arc to the
given point will make the angle desired with the given line.
Problem : By the above Rule (87) show how to construct
a triangle when two angles and the side included by them
are given.
Hint : Lay off a line equal to the given side of the triangle
and at its extremities construct angles equal to
the given angles.
Problem: By the above Rule (87) show how to construct
a triangle when two sides and the included angle are given.
Hint: Lay off an indefinite straight line. With this as a
side construct an angle equal to the given angle.
Then lay off lengths on these sides equal to the
given sides.
56 GENERAL MATHEMATICS
88a. Theorem: Through three points not in a straight line
one circumference and only one can be drawn.
i s
a *** • *•• a •
/
/
/
* •
i
i
/
i
i
' • / ,*" * %
i i . ;> \
i
i
i
i
i
\
\
\
/
/
/
1C
Given: A, 2?, and C are three points not in a straight line.
To Prove : One circumference, and only one, can be drawn
through A y B, and C.
Proof: Draw AB and BC. Construct the perp. bis. of
AB and perp. bis. of BC. Let be the point of intersection
of these bisectors.
Since is in the perp. bis. of AB, the distance from to
A = distance from to B (49).
Since is in the perp. bis. of BC, the distance from to
C = distance from to B.
But it has just been shown that distance from O to B =
distance from to A.
.\ Distance from to A = distance from to 2? = distance
from to C (31).
.*. If is taken as the center and a circumference drawn
with A as a radius, the circumference will pass through
A, B, and C.
Now the center of the circle must be the point of inter
section of the perp. bis. (49).
GENERAL MATHEMATICS 57
But these perp. bis. can have only one point of intersec
tion (8e).
/. is the only point which can be used as a center of p,
circle.
Then, any circumference passing through the three points
must have a radius of AO.
That is: any circumference drawn through A, B, and C
must coincide with the circumference first drawn (16).
That is: one circumference and only one can be drawn
passing through the three given points. Q. E. D.
If the three points were in the same straight line, then the
perpendicular bisector of the line joining the first and second
points would be parallel to the perpendicular bisector of
the line joining the second and third points (60a). Then
since the perpendicular bisector of the lines joining the
points would never meet, there would be no point which
might be taken as the center of the circle passing through
the points — in other words, it is not possible to pass a
circumference through three points in the same line. This
fact may be stated as follows:
88b. A straight line cannot meet the circumference of a
circle in more than two points.
88c. To circumscribe a circle about a triangle {that is, to
Pclss a circumference through the three vertices) . The method
of construction and proof is left to the student (88a).
Exercise
,Let the student prove the following:
1. The perpendicular distances from the center of a circle
to equal chords are equal.
2. A diameter perpendicular to a chord, bisects the chord
and the arc subtended by it.
58 GENERAL MATHEMATICS
89a. Theorem : An inscribed angle is measured by half the
arc intercepted between its sides.
Case 1. When one of the sides of the inscribed angle is
a diameter.
Given: Let the side AB of the inscribed angle CBA pass
through 0, the center of the circle.
To Prove: Z B is measured by Y2 arc AC.
Proof: Draw OC.
OC is the radius of the circle (17)
.\ A OBC is an isosceles A (18, 70)
Z B= Z C (77b)
(I) ZA0C = ZB + ZC (81b)
Substituting for Z C its equal Z B in equation (I)
ZA0C= ZB+ ZB
That is, Z AOC = 2 Z B
1/2 ZA0C= Z5 (33b)
Now ZAOC is measured by arc AC (26)
/. LB is measured by 3^ arc AC.
GENERAL MATHEMATICS 59
Case 2. When the center of the circle is within the in
scribed angle.
g L^T. * J Q
Given: 0, the center of the circle, is within the inscribed
angle ABC.
To Prove: Z ABC is measured by 1/2 arc AC.
Proof : Draw BO, producing it to meet the circumlerence
at some point D.
Since DB is a diameter of the circle,
(I) Z ABD is measured by 1/2 arc AD (Case 1).
(II) Z D5C is measured by 1/2 arc DC (Case 1).
Adding:
Z A5D + Z D5C is measured' by 1/2 arc AD + 1/2
arc DC.
That is, Z A5D + Z DBC is measured by 1/2 (arc AD
+ arc DC).
That is, Z AfiC is measured by 1/2 arc AC (32a).
60
GENERAL MATHEMATICS
Case 3. When the center of the circle is outside of the
inscribed angle.
. 6
Given: Let 0, the center of the circle, be outside of the
inscribed angle ABC.
To Prove: Z ABC is meas. by 1/2 arc AC.
Proof: Draw BO and produce it to meet the circum
ference at D.
Since BD is a diameter,
(I) Z DBC is meas. by 1/2 arc DC (Case 1).
(II) Z DBA is meas. by 1/2 arc DA (Case 1).
Subtracting, Z DBC A DBA is meas. by 1/2 arc DC
1/2 arc DA. That is, Z DBC  Z D5A is meas. by
1/2 (arc DC arc DA).
That is, Z A5C is meas. by 1/2 arc AC. Q. E. D.
89b. A semicircle is half a circle A semicircumference
is half a circumference.
The following theorem may be illustrated by sliding the
arms of a carpenter's square along two nails driven in a
board, while holding a pencil at its vertex. Machinists test
the inside of a hollow casting, which should be semicircular
in crosssection, by sliding a steel square around it; if the
casting is true, the vertex of the square will always touch
the casting, while the arms touch the edges.
GENERAL MATHEMATICS
61
89c. An angle inscribed in a semicircle is a right angle.
6
Given: ACB an Z inscribed in the semicircle A CB of
the O ACBD.
To Prove : Z ACB = 90° or a rt. Z .
Proof: Since ACB is a semicircumference, arc ADB, in
tercepted by AC and CB, is a semicircumference (32a).
That is, arc ADB is 180° (24)
.. Z ACB is measured by 1/2 (180°) (89a)
That is, Z ACB = 90° or a rt. Z. Q. E. D.
The preceding theorem is attributed to Thales of Miletus
(one of the Seven Sages of Greece) who lived about 600 b. c.
At the extremity of a given line to erect a verpendicular to
the line.
A 1 "
6
0/
■^
V
/L
.*'
••••••
Given: GL is a straight line.
62 GENERAL MATHEMATICS
To Construct : A perpendicular to GL at L.
Construction: With any point as (outside of GL, and
not further to the left than the center of the line), as the
center of a circle, and with OL as a radius, draw a circle
intersecting GL at A. Draw AO and produce it to meet the
circle at B.
A straight line drawn from B to L will be ± to GL at L.
The proof is left to the student (89c).
The student may now understand the following:
89d. Rule. At the extremity of a given line to erect a per
pendicular to the line: take any point outside of the given line
(and not beyond its middle point) as a center, and with a
radius equal to the distance from this point to the given ex
tremity, draw a circumference passing through the given
extremity, and intersecting the given line at some point within;
from this point of intersection draw a line through the center
of the circle and produce it to intersect the circumference again.
A straight line drawn from this last point of intersection to the
given extremity, will be the desired perpendicular.
89e. Either part of a circle cut off by a chord is called a
segment. Thus BCDA is a segment of the circle on page
59, and the remainder, enclosed between the arc AB, and
chord AB, is a segment.
Exercise
Let the student prove the following:
1. Angles inscribed in the same segment are equal.
2. Draw a semicircle on stiff cardboard. Attach a small
plumb line to one end of the diameter. Hold the card up
right so that the plumb line intersects the circumference at
some convenient point. A line from the other end of the
diameter to the point of intersection of the chord with the
circumference will be level. Why?
GENERAL MATHEMATICS
63
V. Field Exercise
TO LAY OUT A ROAD IN AN ARC OF A CIRCLE PASSING
THROUGH THREE POINTS
*
Equipment: Two dozen stakes, knotted cord, long cord,
measuring tape, piece of twine, mallet.
Let Ti, T 2 , T z , be three trees (or three buildings or other
objects) past which it is desired to have a road pass in an
arc of a circle.
ftactii Zincs
piecing
Procedure: Having decided on the requisite amount of
clearance (such as two feet) between the trees and the road,
lay off this distance from the trees toward the estimated
position of the center of the circle, marking these distances
with the stakes A, B, and C.
Lay off the lines AB and BC. With the tape find the
middle points of these lines and mark them with the stakes
D and E. Then locate the stake F (see Field Exercise II),
which shall form with D the perpendicular bisector of AB.
Likewise locate the stake <?, which shall form with E the
perpendicular bisector of BC.
Hold the long cord in a straight line past F and D (just
touching them), so that the line DF is sufficiently extended.
Sight from E across G to the point on DF produced, whiqh
64 GENERAL MATHEMATICS
is in the same straight line with E and G. Mark this point
with the stake H , which is the center of the circle.
(Instead of sighting from E to G, another cord may be
used to produce the line EG in the same manner as DF, and
their intersection marked with the staktf H.)
To test the accuracy of the construction, make a loop
in the long cord about H , and stretch the cord tightly to A
to mark the length. Tie a bit of twine about the cord at
the point where it touches A. Then, using the cord about
H as a center, see if B H and CH are the same length as AH.
In case one of the three radii thus tested is longer than
the other, this excess in length may be corrected by moving
the center back that amount along the radius produced.
Thus, if C H is 4 inches longer than A H and B H, move
H 4 inches further from C along the line CH.
For very accurate work, to avoid the error caused by the
varying stretch of the cord, the radii AH, BH and CH may
be laid off with stakes, and the distances measured with the
tape.
When the center of the circle has been definitely located,
the extremities of other radii may be marked with stakes
between A and B } and B and C.
As each radius is laid out, the distance desired for the
width of the road may be measured back along the line of
the radius, and marked with a stake.
This completes the work of staking out arcs of two con
centric circles separated by the width desired for the road,
and with the outer arc passing through the three desired
points A, B } and C.
CHAPTER VI
POLYGONS: USE OF LETTERS TO EXPRESS
GENERAL LAWS
90a. A polygon is a portion of a plane bounded by straight
lines.
90b. The bounding lines are called the sides of the poly
gon, their intersections the vertices, and the angles between
adjacent sides are the angles of the polygon. .
By producing the sides of a polygon, we may form its
exterior angles — but when the angles of a polygon are
spoken of, it means the interior angles.
As we already know, a polygon of three sides is called a
triangle.
90c. The distance around the outside of a polygon is
called its perimeter.
Find the perimeter of this page in inches.
91a. A quadrilateral is a polygon of four sides.
91b. A rectangle is a quadrilateral whose opposite sides
are parallel, and whose angles are right angles.
The surface of this page is a rectangle.
91c. A pentagon is a polygon of five sides.
91d. A hexagon is a polygon of six sides.
92. A diagonal of a polygon is a line joining the vertices
of two angles which are not adjacent. (See AC in the next
figure.)
This definition is not usually applied to triangles, though
it may be — in which case the diagonals coincide with the
sides of a triangle.
65
66 * GENERAL MATHEMATICS
93. Theorem: The sum of the interior angles of a quadri
lateral equals four right angles.
6 — c
Given: A BCD is a quadrilateral.
To Prove: Z i+ Z B+Z C+ Z D=4rt. A.
Proof: Draw the diagonal AC.
This divides ABCD into two triangles.
The sum of the A of the A = the A of the quadrilateral.
The sum of the A of each A = 2 rt. A (80b).
/. The sum of the A of both triangles =2X2 rt. ^ = 4
rt. A.
That is, the sum of the A of the quadrilateral = 4 rt. A
Q. E. D.
94a. Theorem : The sum of the interior angles of a pentagon
equals six right angles.
A <z—
Given: ABCDE is a pentagon.
GENERAL MATHEMATICS 67
To Prove: Z 4+Z B+zC+Z D+Z S=6rt. A.
Proof: From A draw diagonals AC and AD.
This divides ABODE into three triangles. The sum of
the A of the A = the sum of the A of the pentagon.
The sum of the A of each A = 2 rt. A.
.\ The sum of the A of all 3 A = 3 X 2 rt. A = 6 rt. 2f.
That is, the sum of the A of the polygon = 6 rt. A. Q. E. D.
94b. "3X2 rt. A " might be written 3 (2 rt. A).
The symbol ( ) is called a parenthesis.
In the expression 3 (2 rt. A), 3 is the coefficient of the
parenthesis; that is:
94c. A coefficient of a quantity shows how many times
the quantity is taken.
95a. Theorem: The sum of the interior angles of a hexagon
equals eight right angles.
A4f
Given: ABCDEF is a hexagon.
To Prove: ZA+/.B+ZC+ZD+ZE+/.F
= 8 rt. A.
The proof is left to the student.
In like manner, it can be proved that the sum of the in
terior A of a polygon of 7 sides = 10 rt. A, that the sum of
the interior angles of a polygon of 8 sides = 12 rt. A and so
on. All the theorems concerning the sum of the interior
angles of a polygon may be summed up in the following
general one:
68 GENERAL MATHEMATICS
Theorem : The sum of the interior angles of a polygon is
equal to two right angles, taken as many times less two as there
are sides in the polygon.
In stating what we are to prove, we cannot now say " To
prove that Z A + Z B, etc. = 4 rt. A " or " = 6 rt. 4" or
any other particular number of right angles, for we are to
prove the theorem in general; we therefore say in our proof:
" To prove that Z A +Z B + etc. = (n2) 2 rt. A." By
using n, instead of a particular number to represent the
number of sides, we may prove a general rule— thus includ
ing in one statement the rules regarding quadrilaterals,
pentagons, hexagons, and all polygons.
95b. This is the purpose of using letters instead of numbers
in mathematics: to derive general laws.
Note the use of parenthesis in the statement that " the
sum of the angles of any polygon equals (n — 2) 2 rt. A."
The parenthesis indicates that the quantity within it is to
be considered as one quantity — that is, to work out the sum
of the A of any polygon we have first to substitute the proper
number for n, then subtract 2 from it and then multiply
2 rt. A by this number. Thus to find the sum of the angles
of a quadrilateral by this statement, we substitute 4 for n
in the parenthesis and so obtain that the sum of the angles
of a quadrilateral = (4 — 2) 2 rt. A.
Simplifying (4—2), the number inside the parenthesis is 2.
2, then, is the number by which 2 rt. A is to be multiplied.
To find the sum of the interior angles of a pentagon (91c),
we substitute 5 for n in the expression, " (n — 2) 2 rt. A"
and so obtain (5 — 2) 2 rt. A = 6 rt. A.
96a. Any number of quantities enclosed in a parenthesis
must be treated as but one quantity.
96b. If a parenthesis is preceded by a coefficient, each term
within the parenthesis should be multiplied by the coefficient
when the parenthesis is removed.
The reason for this may be seen from the following ex
GENERAL MATHEMATICS
69
ample: Let n = 3 in the expression 2 (n— 2) and we have
2 (3—2) =2 (1) = 2. If we remove the parenthesis before
substituting we have 2n4. Substituting, we have 64
= 2, the same as before.
The proofs just given with regard to the quadrilateral,
pentagon and hexagon, as well as proofs which might be
worked out for all other kinds of polygons, may now, by
the use of the symbols just explained, be all included in the
following general proof:
97. Theorem: The sum of the interior angles of a polygon
is equal to two right angles taken as many times less two
as there are sides in the polygon.
Given: Let ABCDEFG represent a polygon of n sides.
To Prove: Z A+ Z B+ Z C+ etc. = (n2) 2 rt. A.
Proof: From any vertex as F draw all possible diagonals.
The sum of the angles of these A equals the sum of the
angles of the polygon (32a).
Now, in drawing lines from F to all the other vertices, a
triangle would be formed for every side of the polygon ex
cept the two adjacent sides, for evidently lines from F to
G and from F to E would coincide with the sides FG and FE.
That is, there will be two A less than there are sides, that
is, there will be (n2) A.
70 GENERAL MATHEMATICS
Now, the sum of the A of each triangle = 2 rt. A (80b).
The sum of the angles of all the triangles, that is, the sum
of the A of the polygon = (n— 2) 2 rt. A.
In the case of a triangle n = 3 and the quantity inside the
parenthesis peduces to 1, and so we obtain that the sum
of the angles of a triangle equals (1) 2 rt. A = 2 rt. A; that
is, this rule holds for A also.
This rule may be briefly stated as follows : The sum of the
int. A of any polygon = (n  2) 2 rt. A.
Such a brief statement is called a Formula.
98. A formula is a statement of a rule by means of letters
and symbols.
Exekcise
1. Find the sum of the interior angles of a polygon of
seven sides.
2. Find the sum of the interior angles of a polygon of
fifteen sides.
3. Find the sum of the interior angles of a polygon of
eleven sides.
4. How many sides has a polygon if the sum of its angles
equals fourteen right angles?
Solution: 2 (n  2) rt. A = 14 rt. A,
or more briefly
2 (n  2) = 14.
2n4=14.
Since, as stated in the above quotation, when 4 is sub
tracted from 2n we have a number equal to 14, the number
equal to 2n must be 4 greater than 14, or 18. This may also
be obtained by moving 4 from one side of the equation to
the other.
Thus the equation 2n  4 = 14 may be written
2n = 14 + 4
2n = 18
whence n = 9.
GENERAL MATHEMATICS
71
99a. Moving a quantity from one side of an equation to
another is called transposing the quantity. When a quan
tity is transposed it must be changed in sign.
When (4) was taken from the left hand side of the above
equation it was the same as if 4 had been added to that
side of the equation. Suppose that a boy owes $4 at a store.
Since this debt represents money that the boy must pay
out it may be represented as "  4 dollars." Now, if the
boy's father pays the bill at the store, it is the same as if
he had given the boy $4. That is, by removing the (  4)
dollars, he adds a (+ 4) dollars. Hence:
99b. Removing a minus quantity is the same as adding an
equal plus quantity.
100. Either side of an equation is called a member. The
two members of the equation 2n  4 = 14 are 2n — 4 and
14 respectively.
101. An equation may be considered as a balance.
Thus, if equal weights are placed in the pans of the scales
they balance ; if we remove some weight from one side of the
balance, we must remove an equal weight from the other,
in order to preserve the equilibrium (that is to keep one pan
from going up while the other goes down) . Similarly, if we
add a weight to one side of the balance, we must add a
weight to the other; if we double the weight on one side, we
72 GENERAL MATHEMATICS
must double the weight on the other; if we halve the weight
on one side we must halve the weight on the other — in short,
each member must be treated alike, that is, an equation re
mains a true statement of equality if the same quantity is
added to both members, or subtracted from both members, or ij
both members are multiplied or divided by the same quantity.
Exercise
1. How many sides has a polygon if the sum of its in
terior angles equals 20 right angles?
2. What is the number of sides of a polygon if the sum
of its interior angles equals 10 right angles?
3. One number is six times another and the difference of
the two is 540. Find both numbers.
Solution :
Let n = smaller number.
Then 6n = larger number.
From the statement of the problem,
6n  n = 540
That is, 5n = 540
n = 108, smaller number.
6n = 648, larger number.
Check: 648 108 =540
In checking a problem, we substitute the numbers found
in the problem, in place of the unknowns, to see if they give
a true equation.
4. One number is 9 times another, and the sum of the
two numbers equals 80. Find the two numbers.
5. The sum of two numbers is 88. Three times the less
equals twice the greater, plus 29. Find the numbers.
Let n = the smaller number.
Then (since the sum of the two numbers is 88),
88  n = larger number.
From the statement of the problem, 29 must be added to
GENERAL MATHEMATICS 73
twice the greater, to make it equal to three times the less,
that is:
3n = 2(88n) + 29
3n = 1762n+29
Transposing
3n + 2n = 176 + 29
5n = 205
n = 41, smaller number
88— n = 88— 41 = 47, larger number.
Check: 3 (41) = 2 (47) + 29
123 =94+29
123 = 123
6. A rectangle is five times as long as it is wide. If x
represents the width in feet what will represent the length?
What will represent the perimeter?
7. A rectangle is 3 times as long as it is wide, and its
perimeter is 264 ft. Find the length and the width.
8. The perimeter of a rectangle is 868 ft.; and its length
is 6 ft. more than three times the width. Find the length
and the width.
Hint: Let x = width, then 3 x + 6 = length.
9. Find the dimensions of a rectangle whose perimeter is
500 ft., and whose length is 30 ft. less than four times the
width.
10. The perimeter of a rectangle is 664 ft., and its length
is 2 ft. more than twice the width. Find the length and the
width.
11. The perimeter of a rectangle is 160 ft. and its length
is 8 ft. more than twice its width. Find the length and the
width.
12. Find the dimensions of a rectangular field whose
perimeter is 700 ft. and whose length is 50 ft., less than three
times the width.
13. A rectangular field is 5 times as long as it is wide,
and its perimeter is 936 ft. Find the length and the width.
74 GENERAL MATHEMATICS
H. The sum of two numbers is 135. Twice the greater,
minus 25, equals 5 times the less. Find the numbers.
15. The sum of two numbers is 98. Three times the
greater equals 5 times the less, plus 46. Find the numbers.
16. The sum of three consecutive numbers is 39. Find
the numbers.
Note : Since consecutive numbers are numbers which fol
low one another we may represent the numbers as follows:
Let x = one number,
then x + 1 = the second number,
then x + 2 = the third number.
17. The sum of two consecutive numbers is 41. Find the
numbers.
18. The sum of four consecutive numbers is 54. Find the
numbers.
19. Seven times the sum of two consecutive numbers
equals 63. Find the numbers.
20. Three times the sum of three consecutive numbers
equals 72. Find the numbers.
21. Find three consecutive numbers such that twice the
sum of the first and second equals three times the third, plus
three.
22. If three rods are fastened at their extremities so as to
form a triangle, is the figure rigid? (Can the shape of the
figure be readily changed?)
23. If four rods are fastened at their extremities so as to
form a quadrilateral, is the figure rigid?
102. A regular polygon is a polygon which is equiangular
and equilateral — that is, its sides are all equal and its angles
are all equal.
The angle at a vertex of any regular polygon may be
computed by finding the sum of the angles of the polygon
(97) and dividing by the number of sides (since the number
of sides is the same as the number of angles, and the angles
are equal).
GENERAL MATHEMATICS 75
103a. To draw a regular hexagon: Draw a circle; with the
same radius lay off six chords around the circle; these will be
the sides of the regular hexagon.
The proof is left to the student.
Hint: Draw radii from the vertices (25, 77b).
By connecting alternate vertices, an equilateral triangle
(one with three equal sides) is formed. (The proof is left
to the student.) By connecting the three remaining ver
tices another equilateral triangle is formed overlapping the
first. The outer edges of these triangles form a sixpointed
star.
103b. By drawing a circle and in it drawing two diameters
perpendicular to each other and connecting the ends of these
diameters, a square is formed.
(The proof is left to the student.)
By bisecting the arcs thus formed and drawing lines to
their middle points from the vertices of the square, a regular
octagon (figure of eight sides) is formed.
By continuing the process regular figures of 16, 32, etc.,
sides may be formed.
By a similar process with the hexagon, figures of 12, 24,
etc., sides may be formed.
76
GENERAL MATHEMATICS
VI. Shop Exercise
. PLATE PAD
Cut a piece of wood 6" square. Draw diagonals with
pencil. At intersection of diagonals place the point of the
compass and describe a circle touching each of the sides.
With the same center, describe another circle of less diam
eter (choosing a radius which will give a design in good
proportion). In this circle draw a six pointed star (103a).
Follow along the outside of the star with the point of the
pocket knife, pressed into the wood to a depth of about one
sixteenth of an inch. Use any standard woodstain to color
the star. (The scoring with the penknife prevents the stain
from spreading.)
Cut around outer circle with a turning saw; smooth off
with a spokeshave and sand paper.
Other designs may be made in this way by the student.
CHAPTER VII
LITERAL EXPRESSIONS
104. A literal expression is one that makes use of letters
to represent quantities. Thus, the expression for the sum of
the interior angles of a polygon (2n  4) rt. angles is a literal
expression which states a general truth — that is, one that
is true no matter what the value of n is (97) ; or a literal
expression may be used to represent an unknown quantity,
as in the previous exercise, and in the following example.
Exercise
Example : l. Three times a number, plus twice a number,
plus four times the number equals 27. What is the number?
Let x = the number.
Then stating the problem as an equation:
3x + 2x + Ax = 27, /. x = 3
Add
2. 5x + 7x + lis + 2x + %x
3. 3y + 4y + Sy + lly
4. 4z + 2z+ 5z + 102
5. Sy + Qy + 2y + Uy
The above examples, like those in Chapter VI, are ex
amples in Algebra, or Algebraic examples.
105. Algebra is the study of literal expressions.
106. The parts of an example in Algebra which are sep
arated by + and  signs are called the terms.
Thus in: 6a + 36  4d,
6a is a term, 36 is a term, 4d is a term.
77
78 GENERAL MATHEMATICS
107. If an expression consists of only one term it is called
a monomial. Thus, 6d is a monomial.
108. If an expression consists of two terms, it is called a
binomial. Thus, x  y is a binomial.
109. If an expression consists of more than two terms it is
called a polynomial.
110. If we think of the term 6a as being made up of two
factors, 6 and a, either term may be called a coefficient
A coefficient of a term may be any one of its factors
(94c). If one of the factors of a product is a number, while
the rest are letters, we usually mean the number when we
speak of the coefficient, though sometimes we speak of it
as a numerical coefficient. Thus 6 is the coefficient of a in
the term 6a.
111. Any collection of terms is called an expression.
Thus, 6a + 3&  4d is an expression.
We have already learned that the sign ( ) placed around
two or more terms means that they are all to be treated
in the same way; to be worked with as if they were one
quantity. Thus, 2 (3a + 2d + 5c) means that 3a, 2d and
5c are all to be multiplied by 2.
112. The sign [ ] is called a bracket, and is used in the
same manner as a parenthesis.
113. The sign { } is called a brace and is used in the same
manner as a parenthesis.
The terms to be added may also be written beneath each
other. Thus,
Sabc
4afcc
2abc
9abc Ans.
GENERAL MATHEMATICS 79
In the following examples, add only the numerical coeffi
cients :
6.
7x
7. 9yz
8.
\ghm
2x
Zyz
9ghm
5x
«
fyz
6ghm
8x
6yz
&yz
3ghm
hghm
9. 5uvw
10.
• Sxy
%uvw
1.2xy
4cuvw
6.3xy
7uvw
A.Sxy
Suvw
b.Txy
• 2xy
In Arithmetic we learned that if a man earned 4
dollars, and spent 2, and next day earned 3 dollars and
spent 1, we express the dollars to be paid out with a minus
sign in front of them, and what he received with a plus
sign in front of them.
So we may write :
+ 4 dollars + 4d
 2 dollars  2d
+ 3 dollars + 3d
— 1 dollar — Id
+ 4 dollars + 4d Ans.
Since we usually omit the plus sign before a quantity if it
stands alone or at the beginning of an expression, we may
rewrite the example in this manner,
4d
 2d
3d
 Id
4d Ans.
80 GENERAL MATHEMATICS
It is usual to omit the number 1 when it occurs as a
coefficient. Thus, instead of writing "— Id" we may write
simply " d."
Do the following examples, first adding all the plus terms,
then all the minus terms, and subtracting the smaller of the
two sums thus obtained from the larger and giving the re
mainder the sign of the larger term.
11.
U.
3x
12.
4j/2
13,
3ahc
 2x
7yz
9abc
bx
 2yz
 4yz
— 7abc
— 2abc
15.
byz
— limn
16.
Aabc
Gmn
1U
— Smn

Smn
 2ob
4mn
Aran
 7ab
8mn
— Qmn
— 4a6
— 2mn
2x
bmn
— Scde
17. —
18.
—
9x
4cde
—
4x

— bcde
—
6x
7cde
"^
7x
•
bcde
— Scde
114. If a factor occurs more than once in any term, it is
not necessary to write the factor more than once, if we use a
small figure called an exponent, written above and to the
right, to show how many times the letter is taken as a fac
tor. Thus, 7cc may be written 7c 2 , baaa may be written
5a 8 and bbbb may be written fr 4 .
GENERAL MATHEMATICS 81
We learned in Arithmetic that 2 feet + 3 square feet + 6
cubic feet cannot be combined into one term, any more
than we can add bananas and oranges, so in Algebra such
expressions as 2a + 3a 2 + 6a 3 cannot be combined into one
term.
115. Rule: Terms which have different exponents cannot be
added or subtracted.
Terms which have the same letters, each letter having
the same exponent in one term as it has in another, are
called like terms. The numerical coefficient of like terms
need not be the same. Thus, lOab and — Tab are like
terms : so also are 4x 2 yz z and 9x 2 yz z .
116. Rule : Only like terms may be added or subtracted.
Exercise
Add:
1.
5a 2 bc
2.
6cd z
3. \x 2 yz z
2a 2 bc
 3cd 3
— 7xhjz z
7a 2 bc
5cd z
— 2x 2 yz z
6a 2 bc
2cd z
— 3x 2 yz z
8a 2 bc
lied 3
 8x 2 yz z
3a 2 bc
4cd 3
— 5x 2 yz*
4.
ISmnp 2
I
h \lv 2 u z x
— 9mnp 2
\ZvPu z x
4tmnp 2
— 25ti 2 u z x
15mnp 2
\\v 2 u z x
— 7mnp 2
12v 2 u z x
8mnp 2
13v 9 u z x
Parentheses which contain the same quantity may be
combined by combining their numerical coefficients.
Thus:
3 (m + n) + 2 (m + n) + 3 (m + n) = 8 (m + n).
82 GENERAL MATHEMATICS
6 4 (a + 6) 7. 2 (x + y + z) ' 8. 10 (3a  26  c)
5 (a + 6) 3 (x + y + z) 11 (3a  26  c)
6 (a + 6)  5 (x + y + z)  7 (3a  26  c)
3 (a + 6) 6 (x + y + z) 6 (3a  26  c)
9 (a + 6) 4 (x + y + z)  5 (3a  26  c)
 (x + y + z)
We have learned that money to be paid out is expressed
with a minus sign in front of it. Thus, if a man receives
10 dollars and then he receives a bill for 6 dollars (that is,
a notice that he must pay 6 dollars) we may say that — 6
has been added to his 10 dollars. Adding this — 6 dollars
to his 10 dollars means that the man will have 4 dollars left.
In other words, to add a minus number to another number
is the same as subtracting an equal plus quantity from it.
Hence we have the following:
117. Rule for Subtraction: Change the sign of the number
to be subtracted and add.
Exercise
Subtract:
l. 7a 2. 86 3. lid 4. 14m 5. 1 / 4 h 6. .7d
5a 66 _5d _9m VgA JJd
7. 151cd 3 8. 3. Iran 2 9. 43rW 10. 91u*v*x*
49cd 3 2 . 7ran 2 31r 3 s< 4 SluWx*
We learned in Arithmetic that if a man owed 15 dollars,
that is, had — 15 dollars, and had 10 dollars, that is + 10
dollars, the amount he was in debt after payment was 5
dollars, that is, he had — 5 dollars.
So in Algebra we may subtract 15d from lOd by writing
the answer as — 5d. In like manner, any larger quantity
may be subtracted from a smaller by writing the answer
as minus.
GENERAL MATHEMATICS 83
Subtract:
11. ZsH 12. lOafyz 3 13. 21a 3 6c 2 H. Sle 2 fg*
7sH 17 x*yz* 32a 3 6c 2 49e 2 /</
3
15. 39ra 6 n 6 16. .2a6 3 17. n /zeg 2 h b 18. 73d 3 c
72m 6 n 5 7. lab 3 9 / 2 eg 2 ft 5 6 / 9 d*e
If a man owed 20 dollars, and 6 dollars of the debt was
paid for him, he still owes 14 dollars, that is, if from — 20
dollars we subtract — 6 dollars, the answer is — 14 dollars.
So in Algebra, if from — 20a we subtract — 6a the answer
is — 14a. That is, to subtract one minus quantity from
another we follow the Rule just given for subtraction.
Subtract:
19.  13a6 20.  28cd 2 21.  39a 2 6 5 c 6
 bob  lied 2  12a 2 6 5 c 6
22. — 46 3 d 2 23. — 2 /s«/V 24.  lla6 2 c 3
 36 3 d 2  l / h epg*  17a6 2 c 3
26. — 4 (2m + n + p) 26. — 10 (Zx + y — z)
 6 (2m + n + p)  10 (Zx + y  z)
Exercise
SimpUfy the following expressions by combining similar
terms:
1. 4a + 3c + 76 + 2a + 96 + 4c + 26 + a =
2. 2x + 4ty  2 + 3x + 5y + Zz  x + lly =
3. 5x — 3y — z + 3z — lly + 7x — by — Qz =
4. 3mn + 2a6 — 7d — 8c — 3a6 — 4tmn — 2a6 =
5. Sxyz + 2x — 2xyz — 11a; — 4c + lOd — 13# =
6. 3a 3 + 156c + 4c/ 2 + 11a 3  106c + 7a 3 +66c =
84 GENERAL MATHEMATICS
7. 3 (m + n) + 7x + by + 2 (m + n) + 4x + 2y +
3 (wi + n)  2x + 3y + {m + n)
8. 3 [y + z  x] + 2a + 36 + 4 [y + z  x]  3a + 9ft
+ 11 [y + z  x]
118. Rule: To add polynomials, arrange the expressions
one under the other, so that like terms are in columns, and
then add these columns.
Example :
7a + 2b + c; 3c + 2a  6; 5c  9a + 26; 6a + 26  c
Arranged:
7a + 2b + c
2a  6 + 3c
 9a + 26 + 5c
6a + 26  c
6a + 56 + 8c Ans.
Add:
l. 4a + 2c — d 2. Zmn + x + 2y
— 2a + 7c + 5d 5mn — 2x + 7y
— Sa + Qc + 2d  6mn + 3x — Sy
5a + 3c + 9d 7mn + 2y
2a + 8c  6a" 6mn + 2x — 4y
4a + 2c  7d 9mn + 4x — Zy
3a — 5c + 5d 4mn + 2y
3. 2a* + 76c + 3d 4 + ef 4. 9 (re + y) + 7z + 6
4a a + 66c  2d 4 + 6e/ 4 (x + y) + Sz + 2
5a s + 96c + 5d* + Sef 8 (x + y) + 4z + 7
7a 2  56c + 4d 4 + 6e/ 5 (x + y) + 3« + 4
5a 8  46c + 6d 4 + 9 e/  6 (x + y) + llz + 10
8 (x + y) + 52 + 1
3(x + y) + 8«+ 5
GENERAL MATHEMATICS 85
5. 2ran 2  3s 3  2x + 1
2mn 2  2x z + 7x  5
3ran 2 + It? + 4tx + 6
4ran 2 + 5z 3 + 3x + 9
6ran 2 + 4z 3 + 8x + 5
limn 2 + &k 3 + 4s + 1
Perform the following examples in subtraction by chang
ing the sign of the subtrahend (the lower number) and adding
it to the minuend (the upper number) .
1. 7a + 36 + 4c + d 2. 4a + 2b + 3c + /
3a + b + 2c + d 5a + 3b + 6c + /
3.  2m + Sn  2n 2  n 3 4. s 3  3s 2 + 4x + 5
4m + n 2 + 2n 3 s 3 + 2x 2 + x + 4
119. Rule : If a parenthesis, bracket or brace has a minus
sign in front of it } every term within it must have its sign
changed when the parenthesis, bracket or brace is removed.
Thus, simplify:
3a  26  [7a + {2c  (7a  6  c)\  6].
We first remove the bracket, and since it has a minus sign
before it, we change the sign of every number within it.
Thus,
3a  26  7a  { 2c  (7a  6  c) } + 6.
We then remove the next sign of inclosure, the braces, and
as the brace has a minus sign in front of it, we change the
sign of everything within it.
3a  26  7a  2c + (7a  6  c) +6.
Since the parenthesis has a plus sign in front of it no change
of sign will be made when we drop the parenthesis.
3a  26  7a  2c + 7a  6  c + 6.
Combining we have, 3a — 26 — 3c. Answer.
Note that in this example we removed only one bracket, pa
renthesis or brace at a time, and worked from the outside, in.
86 GENERAL MATHEMATICS
Exercise
Simplify:
1. 3a  [26 + 6c] + 6  (2a + c)  a.
2. 7a  [26  {3c + (2a + 46  c)  2c} ].
8.  [2x + x*  {2x + (x*  x 3 ) + 7}  2].
4. y 4  [y 3 + (2y* + Sy)  4]  (y* + 2jf).
5. 3m + 2p + [n — {2m + p — (m + n) + w} J.
6. x 3  [4s 3 + {2x*  (5x + 1) + x 3 }  10].
7.  [6 + a  (c  6)] + 2a + [{ 2a  (3a  56) } + a].
8. {by — fix) — {4x  [5x + (8j/ — 5x) — 4y — 4x] }.
In the following examples, remove the parentheses, leav
ing the brackets, and simplify the results as much as possible
by combining like terms within the brackets.
9 [ (s + y) + *], [(x + v) — z]
10. [4a + (36  5a) ], [4a  (36  5a) ].
11. [ (m  2n) + (3p  q) ] [ (m  2n)  (3p  q) ].
12. [ (5x  2) + (4y  9) ] [ (5x  2)  (4y  9) ].
Vila. Field Exercise
PROFILE OP LAND
The object is to determine the various heights and de
pressions along the line of a proposed road or ditch.
Personnel: One student to read level, one to carry
ranging pole (and later leveling rod), one to drive stakes,
one to keep the record of the measurements made, two to
use the tape.
Equipment: Level, ranging pole, leveling rod, tape, stakes,
mallet.
Procedure : It is first necessary to lay out a straight line
for the ditch to follow. To do this, set up the level at one
end of the proposed ditch and look toward the other end of
it. The student with the ranging pole now advances along
the line of the proposed ditch, seeking the first height or
&
nuutnuiuwti
IMMMNHltl Mil,
I
UNUUUiliUHII
WW0MM V «vV »•
nam
««%%%«•»•«• i
s
V
88 GENERAL MATHEMATICS
depression, where he holds the ranging pole upright. The
student at the level directs him in moving the pole until it
is in his line of sight, by waving his left hand if he wishes
it moved toward his left, his right hand if he wishes it
moved toward his right, and extending both arms horizon
tally when the pole is located correctly. A stake is now
driven in at this spot, and the Recorder marks it A. The
ranging pole is now carried on (VII, upper diagram), its
various positions being marked with stakes, until the entire
line has been staked off. These stakes should be marked
B, C, etc., and the distance from A to B, B to C. etc..
should be measured and recorded.
Since in this part of the work the level is used merely to
lay off a straight line, the water pipe substitute may be used
(as shown in the upper diagram, VII). If so, it must now
be replaced by the water level, or Note at end of Illb
followed.
The leveling rod must now be held upright at each of
the stakes and the heights read off. If the student holding
the leveling rod holds a dark ruler across the face of the
rod and moves it up or down as the observer at the level
directs, it is possible to determine the level at stations so far
distant that the observer could not read the figures. The
readings of the leveling rod at A, B } etc., should be duly
recorded.
Vllb. Drawing Exercise
Rule a line across the paper to serve as a datum line.
Adopt some convenient scale — such as 1 in. =10 ft., and
lay off the horizontal distance measured to scale along the
datum line, thereby locating A, B 9 etc. At these points
draw perpendiculars below the line. Along these perpen
diculars lay off to scale the readings of the level at these
stations. Connect the points thus obtained, thereby show
ing the profile of the land.
GENERAL MATHEMATICS 89
As will be seen, this drawing reproduces the field conditions
to scale — the datum line taking the place of the level line
shown in the lower part of the diagram (VII).
Since variations in the level are usually slight in com
parison with the horizontal distances measured, it is best
to use a larger scale for showing the differences in height.
Thus, if in the drawing the distances between the stakes
were shown on the. scale of 1 in. = 10 ft., it might be ad
visable to use a scale of 1 in. =1 ft. to show the differences
in height — otherwise the profile of the ground as shown in
the drawing might not vary much from a straight line.
Both the horizontal and the vertical scales should be*
marked on the drawing.
Yin
THE CONSTRUCTION AND APPLICATION OF
PARALLEL LINES
Through a given point to construct a line parallel to a given
straight line.
V
#
%
\
n
Given: GL a straight line, P an external point.
To Construct: A line through P, to GL.
Construction: From P draw PM, intersecting GL at M,
any point between G and L.
At P construct Z RPM = Z PJ/L (87).
Produce RP.
RP produced is // to GL (65).
The student may now understand the following:
120. Role : Through a given external point to draw a line par
allel to a given straight line: through the given point draw a line
making any convenient angle with the gii^en line; with this con
struction line as a side and with the given point as a vertex, con
struct an equal angle which (with the angle just drawn) shall
complete a pair of alternate interior angles; the side just drawn
of this last angle, produced through the given point, will be the
desired parallel.
90
GENERAL MATHEMATICS 91
Problem: Let the student show how to construct a line
through a given point, parallel to a given straight line, by
means of exteriorinterior angles (67b).
121a. A parallelogram is a quadrilateral which has its op
posite sides parallel.
121b. If the angles of a parallelogram are right angles,
the figure is called a rectangle. (Compare 91b.)
121c. A square is an equilateral rectangle.
Note : Equilateral means, having all sides of the same
length (102).
122a. Theorem: The opposite sides of a parallelogram are
equal.
Given: A BCD is a parallelogram.
To Prove: BC  AD, and AB = DC.
Proof : Draw the diagonal BD.
In A ABD and BCD,
BD = BD (Identity)
Z ABD = Z BDC (64)
Z ADB = Z DBC (64)
.. A ABD = A DBC (75)
AD = BC (72c)
AB = DC Q. E. D. (72c)
122b. From the above proof it follows that
A diagonal divides a parallelogram into two equal triangles.
Since, if two parallel lines cross two other parallel lines, they
form a parallelogram, it follows from the above theorem th&t *
92 GENERAL MATHEMATICS
122c. Parallel lines comprehended between parallel lines are
equal.
122d. The diagonals of a rectangle bisect each other.
The proof is left to the student.
Note: This theorem is often made use of practically.
For example, when it is desired to locate an electric light or
other fixture in the exact center of the ceiling of a room,
cords are stretched between the diagonally opposite corners
of the room, and their point of intersection is marked as the
center of the ceiling. When " centering" pieces for a lathe
this theorem may be applied in a manner similar to that
used on page 76.
Note: In the following demonstration, the proof that
MN = NO = OP depends upon showing that they are
corresponding parts of the equal A MXN, NYO and OZP.
These A are proved equal by showing that they have 2 A
and the included side of one = 2 A and the included side of
the other. It is here that the difficulty occurs, for, although
Z NMX  Z ONY  Z POZ (being exteriorinterior A
made by the transversal QS cutting the parallels), this
method cannot be used to show that Z MXN = Z N YO,
for M X and N Y are not the same transversal, consequently
the theorem with regard to exteriorinterior angles (67a)
does not apply. Accordingly the angles formed by TB
must be used as the connecting links to show that Z MXN
= Z NYO  Z OZP.
A similar difficulty occurs in proving that MX = NY =
OZ. Let the student not fall into the error of stating that
"MX = NY, since parallels comprehended between paral
lels are equal." This theorem does not apply here; the
pair of parallels which comprehend MX is not the same
pair that comprehend N Y, nor is it the same pair that com
prehends OZ. Here, again, the transversal TR must serve
to connect the parts.
GENERAL MATHEMATICS 93
123. // three or more parallels intercept equal segments on one
transversal they intercept equal segments on every transversal.
T
x I
Given: AB, CD, EF, and GH are //s cut by the trans
versal T R at I, J, X, and L, respectively, so that IJ = J K
= XL; QS, any other transversal, cutting AB, CD, EF, and
GH at M, N, 0, and P, respectively.
To Prove: MN = NO = OP.
Proof: Draw MX, NY, and OZ, // to TR.
Then MX, NY, and OZ are // to each other. (61b)
.. Z ON Y = Z JVMX (67a)
Z POZ = Z ON Y (67a)
(I) .. Z AWX = Z Oi\TF = Z POZ (31)
Again Z MXiV = Z L/X (67a)
Z JV 70  Z JX 7 (67a)
Z OZP = Z XLZ (67a)
But Z L/X = Z/K7 = Z XLZ (67a; 31)
(II) .«. Z AfXiV = Z #70 = Z OZP (31)
Now MX = IJ (122c)
NY = JK (122c)
OZ = XL (122c)
But /J = JX = XL (Given).
(III) .. ilf X = NY =0Z (31).
Then, by I, II, and III,
A NMX = A OiV 7 and A OiV 7 = A POZ (75).
94 GENERAL MATHEMATICS
.\ A NMX  A ON Y  A POZ (31),
.. M N = NO =0P (72c). Q. E. D.
Note : In like manner, the theorem could be proved to
hold for five parallels, or for any number of parallels.
To divide a given straight line into any required number of
equal parts.
*^ — 7 — 7 — 7 6
*C7
Given: AB, a straight line.
To Divide: AB into n (any required number) of equal
parts.
Construction: From A lay off AC, a line of indefinite
length, making any convenient angle with AB.
On AC lay off any convenient unit of length, as u, n
times, beginning at A, and marking the ends of each unit.
From D, the exterior extremity of the nth division, draw
DB.
At each point of division construct a ■// to DB (120).
AB will then be divided into n equal parts.
Proof: At A, construct a // to DB.
Then, since the transversal AD is divided into n equal
parts, the transversal AB is divided into n equal parts
(123).
The student may now understand the following :
124. Rule : To divide a given straight line into any required
number of equal parts: at one extremity of the line draw a
line making any convenient angle with the given line; on this
construction line } beginning at its point of intersection with the
given line, mark off any convenient unit of length as many times
as the number of equal parts into which it is desired to divide
the given line: from the last point thus marked dra/w a straight
GENERAL MATHEMATICS 95
line to the other extremity of the given line: through each point
marked on the construction line draw parallels to the line last
drawn. These parallels will divide the given line into the re
quired number of equal parts.
The vernier is a device for reading very small divisions
of a line, or arc, without the use of a microscope.
 The vernier may therefore be applied to a surveying
instrument to read very small angular divisions (26).
The accuracy of leveling rods (used to find difference in
level) may be increased by being fitted with a vernier
The principle of the vernier may be understood from its
simplest form, such as is used on the barometer. The
barometer is an instrument which measures the atmospheric
pressure by means of a column of mercury which is forced
up into a glass tube by the pressure of the atmosphere.
By the variation of the height of this column of mercury the
air pressure may be determined and the weather predicted.
To measure the height of this column of mercury a scale
(A) graduated in inches may be fixed near the glass tube
containing the mercury. This fixed scale is graduated (that
is, carefully divided) into inches and tenths of an inch, and
the height of the mercury to the first decimal place (that is,
to inches and tenths) may be read directly by comparison
with this scale. A vernier (B) is employed to read the
hundredths of an inch. i
To make the vernier scale, nine of the small divisions on A
are laid off (that is, a line 9/10 of an inch in length is drawn)
and. this line is divided into ten parts, that is, each division
of the vernier is equal to 9/10 *■ 10 = 9/100 of an inch
(while each division of the fixed scale is equal to 1/10 r
10 = 10/100 of an inch). Consequently (since 10/100 
9/100 *= 1/100) each division of the vernier is 1/100 of an
inch smaller than the small divisions of the fixed scale.
In the figure just referred to, the zero mark on the vernier
coincides with the 30in. mark on the fixed scale. The
^
¥
«V
*
Q
/
««>
o
>
•
1
1 1 1
1 1

*)
\l 1 1
 
1 1
1 1

1
1 1/
V **
o
J
\ o
*o o
J
^
¥
5
•
r
V;
1 *
>
,
Q>
I
1 1 1
i i i
•i
\ ~r
1
1 1
i i i
1
1 /
o
/
V *>
J
o
I
(Q
•
r
V
i i i i
^N
•
1
1 1 1
1 1 1 1
\l 1 1

 
1 1
1
'/
\
o
/
\ *>
"J
*3
/
X
GENERAL MATHEMATICS 97
next division of the vernier does not coincide with the next
mark of the fixed scale (the 1/10 mark) but is 1/100 of an
inch below it; the second division on the vernier is 2/100
of an inch below the 2/10 mark on the fixed scale, the third
mark on the vernier is 3/100 of an inch below the 3/10
mark on the fixed scale and so on to the tenth division of
the vernier which is 10/100 below the tenth division on the
fixed scale, that is, the tenth division of the vernier coin
cides with the 9/10 mark on the fixed scale.
Obviously, if we raise the vernier 1/100 of an inch, division
1 on the vernier will coincide with the 1/10 mark on the
fixed scale. Conversely, if, in reading the barometer, we
note that it is the 1/10 mark on the fixed scale which coin
cides with a mark on the vernier, we must add 1/100 to the
height measured directly on the fixed scale and so obtain
30 + 1/100 or 30.01 for the reading of the barometer.
Again, if it is the second division of the vernier which coin
cides with a division on the fixed scale, we must add 2/100 to
the reading on the fixed scale (30) and so obtain 30 + 2/100
or 30.02 for the reading of the barometer; if it is the third
division of the vernier which coincides with a division on
the fixed scale, we add 3/100, and so on.
Keeping these facts in mind, the student is now prepared
to take the readings of the barometer for the positions shown
on the preceding page.
In the first figure, the of the vernier is opposite the
30in. mark on the fixed scale — the reading is therefore 30 . 00.
Suppose the zero mark of the vernier were made to coincide
with the 1/10 mark next following the 30in. mark; the
reading would then be 30 . 10.
In the second figure, the mark of the vernier is between
the 30 in. mark and 30.1 mark on the fixed scale. The
reading from the vernier is obtained by noting that the first
division of the vernier coincides with a division of the fixed
scale. We therefore add . 01 to the reading from the fixed
98 GENERAL MATHEMATICS
scale, 30.00, and obtain 30.01. But if it were the 30.1
mark on the fixed scale that was just below the on the ver
nier, and the first division of the vernier coincided with a
division on the fixed scale, the vernier reading should be
added to 3Q . 10.
From these instances, the student can see that in using
this vernier, the inches and tenths of an inch are read from
the fixed scale, and as many hundredths added to this as the
number of divisions it is necessary to count upwards along
the vernier to find a vernier mark which coincides with a
mark on the fixed scale. In reading the inches and tenths
from the fixed scale, always read to the nearest division on
the scale below the zero mark on the vernier.
The same principle may also be applied to measuring arcs
and, consequently, angles (26). For this purpose the
"fixed scale' ' may be a circular plate and the vernier, marked
on a circular rim which revolves around the fixed circle—
or these positions may be reversed.
Suppose that an arc equal to 29° on the scale is laid off
on the vernier and divided into 30 equal parts; then one
29°
division of the vernier = 29° *■ 30 = — ; changing this
oU
value to minutes,
OQ
^ X 60' = 58'. A division of the scale, 1°, * 60'.
30
Therefore, the difference between a division on the scale
and one on the vernier = 60' — 58' = 2'. This vernier, there
fore, enable^ us to read the angle to the nearest 2'.
Problem: How should a vernier be laid off to read to l'f
to 30"? Dr^w these verniers, remembering that both
fixed scale and vernier arcs must be drawn from the gam?
center. ■< 
The method of reading all the verniers so f ar d^soribe^
may be included in the following; j ov7 / • ^
GENERAL MATHEMATICS 99
125. Rule : To use the vernier: take the reading of the fixed
scale to the nearest division before the zero of the vernier, then
count along the vernier to the division which coincides with a
division on the fixed scale; multiply the number of the divisions
so counted on the vernier by the amount that a space on the
vernier differs from a space on the fixed scale, and add this
product to the reading of the fixed scale.
Note : The vernier is so named after its inventor, Pierre
Vernier, who published a description of it in 1631.
Some verniers, instead of containing one less than the
number of fractional parts in one of the larger divisions of the
fixed scale (as first described) contain one more: for example,
instead of 9/10 as in the barometer vernier just mentioned
such verniers would have a length equal to 11/10 which
would be divided into ten parts.
Although such verniers have the advantage of having their
graduations more distinct (since the spaces between them
are greater) yet the method of reading them is more com
plicated.
Villa. Field Exercise
STAKING OFF PARALLELS
This method is of use (in laying out grounds) to make the edge of a path,
or flower bed, parallel to the front of a building, or other prominent straight,
line. It is useful also in laying out the foundations of a building (II) so that
it may be parallel to a fence or to another building.
D * <
...VJE P
*
"%«,
Given : GL a straight line, P an external point.
To Construct: A line through P parallel to GL.
100 GENERAL MATHEMATICS
Construction : Put a stake at P.
Take any convenient length of tape (as 2n) and swing it
about P as a center until it intersects the given line at A.
Put a stake at A, thereby marking the extremities of the
line PA, of length 2n. Find the middle point of 2n on the
tape stretched along PA and mark it with the stake 5,
thereby bisecting PA.
Put a stake at any convenient point on GL as C.
From C sight across B and place a stake at any convenient
point D, in the same straight line with C and B.
Hold the tape against these stakes, measure the distance
CB and lay off this same distance from B along BD, placing
another stake E at the end of this distance, thereby mak
ing BE  CB. 
Pull up stake D.
PE produced will be parallel to GL.
Proof: In A PBE and CBA
CB = BE (construction)
PB = BA
Z PBE  Z CBA (45)
/. A PBE = A CBA (76)
.\ Z #P4  Z PAC (72c)
Then, since PBA is a straight line (const.) PE produced
is // to GL (65).
Personnel: Leader, stakedriver, two students to use
tape.
Equipment : Tape, mallet, 6 stakes (if distances measured
are longer than the tape, more stakes will be needed) .
Procedure : Drive in a stake near a tree or other land
mark, some 25 yards from the front of a building. By the
above method, stake off a line parallel to the front of the
building through this stake. At a distance suitable for the
width of a path, stake off a line parallel to this.
GENERAL MATHEMATICS
101
VHIb. Pasteboard Vernier.
Make a vernier of pasteboard as in the
sompanying diagram. The "fixed scale"
>uld be graduated in inches and tenths,
i the vernier so graduated as to read to
ndredths (125).
)
front
3/de.
CHAPTER IX
AREAS
126. A Unit of Surface is a square whose sides are one
linear unit in length.
. 127. The Area of a plain figure is the number of square feet,
square inches, or other units of surface which it contains.
Table of Linear Measure
128. 12 inches (in.) = 1 foot (ft.)
3 feet = 1 yard (yd.)
5M yards or 163^ ft. = 1 rod (rd.)
4 rods = 1 chain (ch.)
80 chains (5280 ft.) = 1 mile (mi.)
Note : Feet are often indicated by a short mark placed
after a number and a little above it, inches by two short
marks placed after a number and a little above it; thus,
6' 5" is read, "six feet five inches."
129. For measuring area, the units most in use in the
United States are contained in the following:
Table of Square Measure
144 square inches (sq. in.) = 1 square foot (sq. ft.)
9 square feet = 1 square yard (sq. yd.)
For the measurement of larger areas:
10 square chains = 1 acre (A.)
640 acres = 1 square mile (sq. mi.)
102
GENERAL MATHEMATICS
103
After studying the following theorem (131) let the student
compute the number of yards in a square rod, the number
of rods in a square chain, and the number of square feet in
an acre.
130. Plain figures that have equal areas but cannot be
made to coincide are called Equivalent.
Thus, a circle with an area of one square foot is equivalent
to a square which is one square foot in area.
131. Theorem: The area of a rectangle is equal to the
product of its base by its altitude.
i
•
•
•
•
■
i
4
«
L.
!...._.
T' C
Given : RECT is a rectangle, m units in length and n units
in width.
To Prove: Area of RECT = m X n.
Proof : At the end of each unit of length in the base TC,
draw JLs to TC. These lines will be _L to RE also (62).
These JLs will be parallel to each other (60a).
Then E'ECC, R'E'C'T 1 , etc., are all rectangles (121b).
Each unit of length in TC is a side of a small rectangle,
that is, there are as many small rectangles as there are
units of length in TC, or m small rectangles.
At the end of each unit of length in EC, draw JL s to EC.
These lines are JL to E f C also (62).
Then E'ECC is divided into n rectangles.
These rectangles are squares (121c).
104
GENERAL MATHEMATICS
There are as many of these squares as there are units of
length in EC.
That is, there are n squares in the rect. E'ECC.
By prolonging the Is to UT' in the next rectangle,
R'E'C'T' may be divided into n squares.
Continuing the process each of the small rectangles may
be divided into n squares.
There are m of these small rectangles, and each is divided
into n squares,
/. the whole figure is divided into m X n squares.
But each square is a unit of surface (126).
.*. the number of units of surface in RECT (that is, the
area of RECT) = m X n. Q. E. D.
Note : In the above proof, m and n are used as the length
and width in order that the proof may be general (95b).
132. Theorem : The area of a parallelogram is equal to the
product of its base times its altitude.
tl
Given: ML is the altitude of the parallelogram PARI
To prove: Area PARL = ML X PL.
Proof: Produce RA; from P draw PF } _L to RA pro
duced.
ML is // to PF (60a).
In A FPA and MLR
Z MRL  Z FAP (67a)
GENERAL MATHEMATICS
105
I.
II.
III.
By I, II, HI,
Z LMR  Z PFA (37b)
Z RLM  Z APF (80c)
22L = AP (122c)
LM = PF (122c)
A RLM = A APF (76).
Now, if to the figure PAML we add the triangle LMR
we obtain the parallelogram PARL, and if to the same figure
PAML we add the triangle AFP (the equal of RML) we
obtain the rectangle PFML.
.'. area of the rectangle = area of the parallelogram (34b).
This might be expressed by the following equations:
AMLP + LMR = PARL.
AMLP + PFA = PFML.
A LMR = A PFA (already proved).
AMLP = AMLP (Identity).
.. PARL  PFML (34b) as stated above.
Now the area of FMLP = PL X ML (131).
Substituting for area of FMLP its equal, area of PARL
we have:
Area of PARL = PL X ML. Q. E. D.
133a. Theorem: 77ie area of a triangle is equal to half the
product of its base by its altitude.
Given: CD the altitude of A ABC, AB the base.
To prove: Area of A ACB = 1/2 AB X CD.
106
GENERAL MATHEMATICS
Proof: Draw BE // to AC, CE // to AB.
ACEB is a parallelogram (121a).
/. CB is a diagonal of the parallelogram.
.*. CB divides ACEB into two equal parts.
That is, A ACB is 1/2 parallelogram ACEB (122b).
But area of ACEB = CD X AB (132).
/. Area of A ACB = 1/2 CD X AB. Q. E. D.
133b. This may be briefly stated as area A = %b X a»
Since the order of factors is immaterial (that is, since
2 X 3 X 5 is the same as 3 X 5 X 2 or 2 X 5 X 3) the
above may be written area A = a X Yi b.
That is,
The area of a triangle equals the product of the altitude
times onehalf the base.
133c. If two sides of a quadrilateral are parallel, and
two sides are not parallel, the figure is a trapezoid. The
parallel sides of the trapezoid are called its bases.
133d. The area of a trapezoid equals the product of tk
altitude times onehalf the sum of the bases.
Given: AD and BC are the bases of the trapezoid
A BCD; BF its altitude.
To Prove: Area ABCD = BF X h (AD + BC).
Proof : Draw diagonal BD; produce BC and from D draw
a J_ meeting BC produced at E. The area ABCD = area
A ABD + area A BCD (32a).
GENERAL MATHEMATICS
107
I. Area A ABD = BF X h AD (133a).
Area A BCD = EDX\BC (133a).
But ED = BF (62, 60a, 122c).
II. .. Area A BCD = BF X \ BC. Why?
Adding I and II,
Area A ABD + area A BCD = BF X h AD + BF
X\BC.
Or, since both \ AD and \ BC are multiplied by BF,
Area A ABD + area A BCD = BF X (I AD + \ BC).
Again, since both AD and BC are multiplied by \,
Area A ABD + area A BCD = BF X h (AD + BC).
That is, area ABCD = BFX i (AD + BC). Q. E. D.
Exercise
l. The area of a square is 36 feet. How many feet in each
side?
X
Solution :
The side of the square may be represented by x.
Then the area equals x X x (131).
That is, £ X # = 36.
To find x y evidently we have to find a number, which,
multiplied by itself, equals 36. This number, we know from
Arithmetic, is 6. That is, x = 6. Ans.
In the above solution the expression, "x X x" might
have been written "x 2 " which is read "x square" (114).
108 GENERAL MATHEMATICS
The solution of the above example might therefore have
been written in this form:
1. x 2  36.
II. x = 6. Ans.
The second equation is obtained from the first by extract
ing the square root of both sides.
From the diagram of this problem, in which the area of the
square is marked 36 and the side of the square x (from which
we obtained that x = 6) we may state the method by which
we obtain (II) from (I), in the following form:
134a. Rule : // one quantity is equal to a second, the square
root of the first equals the square root of the second.
Or,
134b. Rule : Extracting the square root of both sides of a
true equation gives a true equation.
2. The area of a square tile is 81 in. How long is one side
of the tile? What is the perimeter of the tile?
3. Two equal square tiles, placed side by side, cover an
area of 50 sq. in. Find the length of the side of one of the
tiles.
Solution: 2x* = 50
x*  25 (101)
x = 5
4. Two square tiles, placed side by side, cover an area of
128 sq. in. Find the length of the side of one of the tiles.
5. Twice the area of a certain square is 338 square in
Find the length of the side of the square.
6. How many square yards of surface are there in a floor
21 feet long and 15 feet wide?
7. How many bricks will be required to build a sidewalk
6' wide and 72' long if the bricks are so placed that the face
of the bricks 9" long and 4" wide is uppermost?
8. How many acres are there in a rectangular field 13
chains long and 1\ chains wide?
GENERAL MATHEMATICS
109
Hint: Since there are 10 sq. chains in an A., to find the
number of acres in a given number of square chains, the
decimal point is moved one place to the left.
9. How many acres are there in a rectangular field 31 rods
long and 24 rods wide?
10. The altitude of a triangle is 2 . 78 chains and its base is
2 . 75 chains. What is its area expressed as a decimal of an
acre?
11. A courtyard is in the form of a trapezoid, one base of
which is 24 feet and the other 32 feet. The perpendicular
distance between these bases is 25 feet. Find the area of
the courtyard.
12. To find the area of a field it is divided into three triangles.
The altitude of the first triangle is 5.48 ch., and its base
6.78 ch.; the altitude of the second is 6.21 ch. and its base
8.78 ch.; the altitude of the third is 7.21 ch. and its base
6 . 78 ch. Find the area of the field in acres.
IX. Field Exercise
FINDING AREA OF FIELD BY BASE AND ALTITUDE METHOD
Equipment: Leveling instrument; ranging pole; knotted
cord; tape; dozen stakes.
Procedure: The field is to be divided into triangles as
x
AMaahMBHaaMOM
"7*
ft
¥■
—Hr
110 GENERAL MATHEMATICS
nearly equilateral as possible (as this is the best form for
computation).
Start at one corner of the field as A and with the leveling
instrument lay off a line intersecting the other side as at B,
a perpendicular from C to AB will be the altitude of the
triangle. The remainder of the field may be divided into
triangles in the same way and their altitudes constructed.
In constructing the altitude of a triangle, as ACB, one
of two methods may be used.
If the perpendicular distance from C to AB is less than the
length of the knotted cord (or tape) the following method
may be used.
Hold one end of the cord at C and describe two arcs
cutting AB } at D and E } which should be marked with
stakes.
(To aid in finding the exact point where the cord crosses
the line A B y the tape should be stretched along the stakes.)
Measure the distance between D and E, and at the
middle point drive a stake F.
CF will be perpendicular to AB at F (since C and F are
each equidistant from D and E).
In case the distance of B from A and G is too long for the
cord or tape to be used to construct a perpendicular from
B to AG, the following method may be employed.
Estimate as closely as possible the foot of the perpendicu
lar from B to AG and mark it with a stake as H . At H erect
a perpendicular to AG as JH. This will probably not go
through B, but at some distance to one side of this point.
Measure the perpendicular distance from B to this line,
as JB. Lay off the distance BJ from H and mark this dis
tance with a stake L. A perpendicular erected at L will
pass through B.
The bases and altitudes of the various triangles being
measured, their areas may be found and these added to
find the total area of the field.
CHAPTER X
MULTIPLICATION OF LITERAL EXPRESSIONS:
FACTORING
It has already been explained in Chapter VI that such
an expression as 5 (a + 6) means that (a + 6) must be taken
5 times. Accordingly, if we remove the parenthesis, each
term within it must be multiplied by the coefficient in front
of it. Thus,
5 ( a + 6) = 5a + 56
Similarly, — 5 (a + b) = — 5a — 56
and x (y + 2) = xy + 2x
It has already been explained that subtracting a minus
quantity is the same as adding an equal plus quantity,
that is, as explained in Chapter VI, — (— 4) = + 4. Further
more, if we subtract a minus quantity twice it is the same as
if we had added double the plus quantity, that is, — 2 (— 4)
= + 8. In the same way, — 3 (— 4) = 12 and so on. All of
these results may be included in the general statement,
— a (— 6) = a&,
where a has any value, and 6 any value, that is,
135. Rule: The product of a minus quantity by a minus
quantity is a plus quantity.
Note: This is stated more briefly, ({ minus times minus
gives plus."
Note: As the student already knows, u minus times plus
gives minus"
Accordingly,— 4 (a— 6) = — 4a + 46.
ill
112 GENERAL MATHEMATICS
Exercise
Remove Parentheses:
1. 5 (2a + 36 + 4c) 5.  6 (46  3c + d)
2. 3 (6a 2  7a + 2) 6.4 (8a  76*  2c)
3.2 (3s + 4j/  2z) 7.  8 (6*  6 + 1)
4.5 (2a  36 c) 8. 13 (z 2 + x + Sy  4)
As has been already explained (114) a 2 means a X a and
a 3 means a X a X a and so on, the exponent showing how
many times the letter called the base is taken. Accordingly,
a 2 X a = a 8
In like manner,
(a&) X (a&) 2 = (a&) 8 ;
since (a&) 8  (db) X (a&) X (a&) =
aXaXaXbXbXb = a 3 & 8 .
That is, (a&) X (a&) 2 = a s & 8 .
Similarly, (2a&) X (36c) X (2ac) = 12a 2 6 2 c 2 .
Applying terms previously explained, the above might be
called a problem in multiplication of monomials of the same
base ; and the method used stated in the following :
136a. Rule: To multiply monomials having the same base;
add the exponents and write this sum as the exponent of the base,
and write the product of the numerical coefficients as the co
efficient in the answer.
In multiplying monomials which have a common factor
but which are not the same throughout, the above rule may
be applied to the factors which are common to different
terms, thus,
(a 2 &) X (36 2 ) X (4a6c) = 12a 3 6 4 c
The method used in the above problems may be stated in
the following :
136b. Rule: To multiply monomials, write the product
of all the numerical coefficients as the coefficient in the answer
and after it write each literal factor with the exponent which
GENEBAL MATHEMATICS 113
is formed by adding together the exponents of this letter in all
the quantities to be multiplied.
This rule may be applied in the following examples:
Simplify:
6a (ab  2a 2 + ac)
Solution :
6a (ab  2a} + ac)  6a 2 6  12a 8 + 6a*c
Simplify :
 3a (2a6  4a6 2 + 7a 2 )
Solution :
 3a (2a6  4a6 2 + 7a 2 ) 6a 2 6 + 12a 2 6 2  21a 8
Remove Parentheses in the following :
9. 26 (6  2c  d) 13. 7xy (2x + 3y  5)
10. 3a (2a  26  4c) 14. 9x 2 (3y  5z + 6x)
11. 4a6 (a6  26 + 3a 2 ) 15. 3a6 (26  3a6 '+ 6 2 )
12. — 2mn (n 2 — 2m + 3n) 16. a 2 (a + a6 — 3c)
Multiplication may also be expressed thus,
3a6 + 4a + 1
2
This indicates that each term of the upper quantity is to
be multiplied by 2. The answer, therefore, is the product,
6a6 + 8a + 2, which should be written below the line.
Multiply:
17. x 2 + Zx + 5 20. a 2 + 2a6 + 6 2
x ab
18. 7a? + 2x + 3y  5 21. 3x 2 + 2xy +Sy 2  2x + Sy  6
2x Sy
19. a 2 — 6 + h % — c 2 22. mn + n 2 — m 2 + 3m + n
36 2 2ran
114 GENERAL MATHEMATICS
The same method may be used when the multiplier is a
binomial, thus,
We first multiply each term in the multipli
2 X + 3 cand by z and write this result as the first
x \ \ partial product, then we multiply the mul
9 t o tiplicand by 1, and write this result as the
x 9 X , o second partial product, being, careful to
x keep similar terms under each other, and if
2x* + 5x + 3 there is no similar term in the first partial
product, we place the result to one side as
in a column by itself. Thus, writing the second partial
product, we place 2x under 3x, but since there is no term
free from x in the first partial product, it is necessary to
place the 3 to one side. After the partial products are
written down in their proper columns, we draw a line and
add the columns. The same method may be used when
the multiplier is a trinomial, thus,
m 2 + Tnn + wi
3m+ 2n + 4
3m 3 + 3m 2 n + 3m 2
+ 2ra 2 n + 2mn 2 + 2mn
+ Am 1 + 4mn + 4m
3m 3 + 5m 2 n + 7m 2 + 2ran 2 + 6ran + 4m
Rearranging, 3m 3 + 7m 2 + 5m 2 n + 2mn 2 + 6mn + 4m
The work of multiplication may often times be made easier
by first arranging the terms in some definite order, generally
according to descending powers of some letter. Thus, con
sider the following problem in multiplication:
, I 1 I o If the multiplication is attempted with the
J° T k _i q 2 terms arranged in this haphazard fashion it
x — — will be found much more confusing than
when the work is arranged in this manner:
x 2 + 3x + 1 •
3x* + 2x + 5
GENERAL MATHEMATICS 115
This method may be generalized for any number of terms
as follows:
137. Rule: To multiply polynomials: — both multiplicand
and multiplier having been arranged in descending powers of
the same letter, multiply the multiplicand by each term in the
multiplier, keeping like terms in the same column, and add the
partial products thus obtained.
Exercise
Multiply:
l. 2x + 3j/ 2. 2a + 3 3. 4j/ + 5
2x+ y a + 2 3y + 2
4. 4x — 3 6. 6  5x 6. 2x — 7
x — 5 x — 2 7 — x
7. 3x  1 8. X 2 + X + 1 9. X 2 * X + 1
bx — 2 x + 1 x — 1
10. a 2 — 3a6 + b 11. a 2 + 3a  26 12. 3m 2 + 2mn + n 2
a + b a — 36 m — 2n
13. 2ra 2 + 3m + 1 14. a + 6 + c 15. a 2 + 26 + 5
m 2 + 2m + 1 a  36 + 1 a + 6 3
138. Multiplication may be expressed by writing two or
more quantities, enclosed in a parenthesis, in the same line,
with no sign between them.
Thus, (2a) (36) means that 2a is to be multiplied by 36,
and (2a + 6 + c) (3a +46) (a — 6 + 2c) means that 2a +
b + c is to be multiplied by 3a + 46 and then this product
is to be multiplied by a — 6 + 2c.
116 GENERAL MATHEMATICS
139. To expand means to carry out a process which is
indicated. Thus, "Expand (3a + 6) (2a — &)" means
to multiply 3a + b by 2a — b.
Expand
16. (x 2 + xy + y 2 ) (x  y)
17. (x 2  xy + y 2 ) (a; + y)
18. (a 2  2a + 4) (a + 2)
19. (x 2  3x + 9) (a; + 3)
As was learned in Arithmetic, the square of a given
quantity is that product obtained by multiplying a quantity
by itself.
The squares of certain algebraic expressions are so im
portant that they deserve special attention.
If we expand (a + ft) 2 , that is, if we multiply (a + 6) by
(a + 6) we obtain the product a 2 + 2a6 + 6 2 . This result
might be stated in words instead of letters, in the following:
140. Rule : The square of the sum of two quantities is equal
to the square of the first, plus twice the product of the first by
the second, plus the square of the last
Expand (2x + 3t/) 2 by the above rule.
Solution :
The square of (2x + 3t/) 2 = the square of the first term,
(2x) (2x), or 4x 2 , plus twice the product of the first term by
the second, that is, 2 (2x) (3y) or 12xy, plus the square of the
last term (Sy) (3y) or 9j/ 2 . The complete answer is, then:
(2x + Sy) 2 = 4z 2 + 12xy + 9j/ 2 .
Expand the following by the above rule.
20. (x + l) 2 21. (y + z) 2 22. (m + n)*
23. (2x + l) 2 24. (Sy + l) 2 26. (a + 26) 2
26. (3m + 2n) 2 27. (3a + 26) 2 28. (Sy + 4z) 2
t 29. (4c + l) 2 30. (a + 5) 2 31. (4c + 3) 2
32. (6 + 7h) 2 33. (8 + 9m) 2 34. (9 + 5a)*
GENERAL MATHEMATICS 117
The pfoove rule, of course, applies as well to numerical
quantities as to literal quantities. By its application num
bers may be squared mentally; thus,
Square 34.
Solution :
(34) 2  (30 + 4)*  900 + 2 X 4 X 30 + 16
 900 + 240 + 16
 1140 + 16
= 1156 Ans.
As will be seen from the above, to apply the rule so
that numbers may be squared mentally, the number is
separated into parts, one of these parts being the next lower
multiple of 10— as here, the number is separated into 30
and the remainder.
Note: In finding the middle term, it is easier (in mental work) to double
the second quantity before multiplying by the first — rather than multiplying
the two together and then multiplying by 2.
Square the following numbers mentally by the above
method:
36. (a) (b) (c) (d) (e) (f)
31 43 52 73 84 22
(g)
(h)
(i)
(j)
(k)
(1)
62
71
63
81
74
33
By means of the theorem which states that the area of a
rectangle is equal to the product of the base by the altitude
we may prove by a geometrical construction the truth of
the above rule (140) and consequently of the formula of
which it is the verbal expression.
To Prove: (a + ft) 2 = a 2 + 2ab + b 2
Let AB be a, line a units in length, and BC be a line b
units in length. Then the line AC is (a + 6) units in
length.
118
GENERAL MATHEMATICS
Construct a square with AC, that is, (a + 6), as a side, and
within this square construct a square with AB, that is, a,
as a side, and produce the interior sides of the smaller square
to meet the sides of the larger square; thus,
K
h
(I) Since AC is (a + b) units in length, the area of
ACGK contains (a + &) 2 units (131).
(II) This square ACGK is made up of the square ABED,
the square EFGH and the rectangles BCFE and DEHK.
(III) The square ABED (since the side AB is a units in
length) is a 2 units in area (131). The square EFGH (the
side of which EF = BC = b units in length) contains 6*
units in area.
The rectangle DEHK = the rectangle BCFE.
(For DE = BE, being sides of the same square, and
EF = EH, being sides of the same square.)
That is rect. DEHK + rect. BCFE = 2 rect. BCFE.
But the area of BCFE = BE X BC = a X b (131).
(IV) /. 2 rect. BCFE = 2ab.
Then, by I, II, III, and IV,
(a + by = a 2 + 2ab + b\ Q. E. D.
GENERAL MATHEMATICS 119
Expand
36. (a + 6 + .c) 2 39. (x + y + z)*
37. (3a + 2b + c) 2 40. \m + n + 2p)»
38. (m + 2n + 3p) 2 41. (2x + Sy + 4*) 1
Another important square is that of the difference of two
quantities. This may be expressed by the formula, obtained
by actual multiplication,
(a  b) 2 = a 2  2ab + 6 2
That is, since a may have any value assigned to it, and 6
may have any value assigned to it, this formula is the literal
expression of the following :
141. Rule: The square of the difference of two quantities
is equal to the square of the first, minus twice the product of the
first and second, plus the square of the last.
The only difference, then, between the square of the sum
of two quantities and the square of the difference of two
quantities, is that in the first case the sign of the middle term
of the answer is plus, while in the second case the middle
tennis minus. ,
Expand
Exercise
1. (x — y) 1 2. (m — n)*
8. (X  l) 2
4. (2a — b)* 6. (3w  n) 2
6. (2x  3y)*
7. (o  5) 2 8. (5x  3y) 2
9. (3a  76)»
10. (5ro — 7n) J 11. (An — 3d) 2
12. (5c — 4n) 2
13. (66  c) 2 14. (5d  2c) 2 •
16. (9n  l) 2
16. (x  10) 2 17. (x  12) 2
18. (2y  5) 2
The above rule may be used to square
numbers mentally.
Square 37.
Solution:
r*
(37) 2 = (40  3) 2 =1600  6 X 40 + 9
= 1600 240 + 9
= 1360 + 9
'
■  1369.
/
120 GENERAL MATHEMATICS
Square the following numbers mentally:
19 (a) (b) (c) . (d) (e)
18 29 47 58 67
(0 (g) <M (i) Q)
38 46 89 78 36
When a student is called upon to square a number men
tally he may use either of the formulas, (a + ft) 2 = a 2 + 2al
+ 6 2 , or, (a  ft) 2 = a 2  26b + 6 2 . The student should
choose the formula which makes 6 as small as possible.
Thus, in squaring 48 we should think of the number as
(50 — 2) thereby using the (a — 6) 2 formula, rather than
thinking of the number as (40 + 8) and so using the (a + 6) 2
formula. (That is, we should use the rule of art. 141, rather
than the rule of 140.) The reason for the choice of a formula
is obvious in the above example, since 2 is easier to use as
a multiplier than 8.
Choose the more convenient of the preceding formulas,
and square the following numbers mentally:
20. (a) (b) (c) (d) (e) (f) (g)
83 36 69 21 . 57 73 86
By means of the theorem which states that the area of
a rectangle is equal to the product of the base by the altitude,
we may prove by a geometrical construction the truth of
the above rule (141), and consequently of the formula of
which it is the verbal expression.
To Prove: ( a 6) 2 = a 2  2ab + b*
Given the line AB, a units in length, AC, 6 units in
length, and CB, their difference, which is, therefore, (a — 6)
units in length.
The above formula may be written:
(a  6) 2 = a 2 + & 2  2a6
GENERAL MATHEMATICS
121
This brings out more clearly that twice the product of a
ind b is subtracted from the sum of the square of a and the
square of b.
The geometrical proof consists in showing that the square
>f which the side is (a — 6) is equal to the remainder left
>y subtracting twice the rectangle, the altitude of which is
<. and the base of which is 6, from the area of the polygon
A
H
di
ab
£
b
C
n
ab
'6
Eormed by adding the square whose side is a to the square
whose side is 6, that is, in showing that
(a  by = a 2  2ab + b*
Upon AB (a units in length) , construct a square, ABFG,
the area of which therefore contains a 2 units.
Upon AC (b units in length), construct an exterior square,
HKCAj the area of which therefore contains 6 2 units.
(I) The area of the entire polygon (that is, BFGHKC)
is therefore equal to a 2 + b 2 .
(II) On the side CB (a — b) units in length, construct a
square, CBED, the area of which is therefore (a — 6) 2 units
in length.
122 GENERAL MATHEMATICS
Produce ED to L.
EF = AC (since each is obtained by subtracting a side
of the square CBED from a side of the square ABFG).
GF  LH [GF =a, and LH  GH  GL = a + 6  & = a]
.\ Rect. GF#L = LDifff .
(Ill) That is, area HKDEFG = 2 area GFEL = 2o6.
Then, by I, II, and III, if from the area of the whole figure
(a 2 + & 2 ) we subtract the polygon KDEFGH, the area of
which is 2a&, we have left the square CBED, the area of
which is (a — &) 2 , that is
a 2 + & 2  2ab  (a  &)*
Or by transposition
(a  &)* = a*  2a& + &*. Q. #. Z).
If the multiplier or multiplicand contains parentheses, it
is possible to multiply them before removing the parentheses,
and is sometimes more convenient to do so. Thus,
(a + b) + c
2 (a + b) + 3c
2 (a + 6) 2 + 2c (a + b)
. 3c (a + b) + 3c 2
2' (a + b) 2 + 5c (a + b) + 3c 2 We may afterwards re
move the parentheses, thus,
2 (a 2 + 2a& + & 2 ) + 5ac + 56c + 3c 2 =
2a 2 + 4a& + 26 2 + 5ac + 56c + 3c 2
Multiply the quantities in the brackets before removing
parentheses, then remove the parentheses, and combine:
21. [ (a + 6)  2d] [3 (a + 6) + 4d]
22. [5 (a + c)  Id] [4 (a + c)  3d]
23. [ (m + n)  5] [2 (m + n)  7]
Expand the following before removing the parentheses,
then remove the parentheses and combine:
24. [ (a  2) + 6] 2
GENERAL MATHEMATICS
123
Hint: This means multiply (a — 2) + 6 by (a — 2) + 6
as in the above examples.
26. [ (x  y) + 5] 2 27. [ (a  36) + 2c] 2
26. [ (2x + y)  3] 2 28. [ (a  3) — 2d] 2
These problems may be expanded by the preceding for
mulas, using 140, if the sign between the parts is +, or 141,
if , thus:
[ (a + 6) + 2c] 2 = (a + 6) 2 + 4c (a + b) + 4c 2
 a 2 + 2ab + b 2 + 4ac + 46c + 4c 2
Again
[ (a  6)  3c] 2 = (a  6) 2  6c (a  6) + 9c 2
 a 2  2a6 + 6 2  6ac + 6c6 + 9c 2
Choose the proper rule (either 140 or 141) and expand:
; (a + 6) + 3] 2 32. [ (a + 6)  (c + d) ] 2
(a + 6)  4] 2 33. [ (m + 5)  (n  n) ] 2
; (6  c)  d] 2 34. [ (2x + y)  (a + 56) ] 2
Another important product is that obtained by multiply
ing the difference of two quantities by their sum, represented
by (a  6) (a + 6)
Multiplying: ^+6
 6
29.
30.
31.
a
a
a 2 + ab
ab  6 2
a s
 6 2
Since a may have any value assigned to it, and 6 any
value assigned to it, the above result may be stated as the
following :
142a. Rule: The product of the sum and difference of two
quantities is equal to the difference of their squares.
Thus: Expand (2x+ Sy) (2x  3y)
Following the above rule, we square the first term (2z)
and obtain 4z 2 , then square the last term (Sy) and obtain
9y 2 , from the first square we subtract the second, obtaining
the answer, 4# 2 — 9y 2 .
124 GENERAL MATHEMATICS
Exercise
rule:
2.
4.
2n) 6.
8.
(m + n) (m — n)
(a + 3) (a  3)
(2a + 3) (2a  3)
(3y + 2z) (3y  2zj
Expand by the above rule:
1 (z + 2/) (s  2/)
3. (c + d) (c — d)
6. (3m  2n) (3m + 2n)
7. (2jc + y) (2x  y)
9. (5m + 3n) (5m  3n) 10. (6  82) (6 + 82)
By the above rule we may multiply certain numbers
mentally; thus:
Multiply 37 by 43.
Solution :
(37) (43)  (40  3) (40 + 3)
= 1600  9 = 1591. Ans.
Note that in order to multiply two numbers together by
the above rule, one of them must be a certain small quantity
greater than a number of which the square can be readily
found, while the other is the same amount less than the
same number. This number need not necessarily be a mul
tiple of 10; for example,
Multiply 13 by 11.
Solution:
(13) (11) = (12 + 1) (12  1) = 144  1 = 143.
Any number of which the square is known may be made
the a in the (a + b) (a — b) formula.
Multiply the following by the above rule :
11.
(a)
(18) (22)
(b)
(32) (28)
(c)
(33) (27)
(d)
(44) (36)
(e)
(51) (49)
(0
(63) (57)
(g)
(82) (78)
GO
(67) (73)
(i)
(48) (52)
GENERAL MATHEMATICS
125
By means of a geometrical construction we may prove
the truth of the above rule, as expressed by the formula.
To Prove: a 2  b 2 = (a + b) (a  6)
Given the line AB, a units in length, the line CB, b units
in length.
On AB construct a square, the area of which will there
fore contain a 2 units; on CB construct a square, the area of
which will therefore contain b 2 units.
■ a
H
r :
G
b
b
*
a
F
If from the area of the large square, a 2 , we subtract the
small square b 2 , we have left the polygon ACGFDE; in
other words,
(I) a 2  b 2 = ACGFDE.
It now remains to be shown that
the area of ACGFDE = (a + b) (a  6)
HG = HC  CG, that is, HG = (a  6)
EH = ED  HD, that is, EH = (a  6)
.*. if EA and C// are produced to / and K, respectively,
so that EI and HK wUl equal GF (or 6), the rectangle
126 GENERAL MATHEMATICS
EIKH will equal the rectangle HDFG (having equal bases
and equal altitudes).
Now AIKC = EHCA + IKHE
and EDFGCA = EHCA + HDFG
(II) .. IKCA = EDFGCA (34b).
Since EI = b and EA = a, the side AI = (o + 6) ;
AC = (o  6) (Since A£ = a, and C5 = b)
(III) .. /XCA  (a + b) (a  6)
.. by (I), (II) and (III),
a*  b* = (o + 6) (a  6) Q. #. D.
This rule may also be applied to more complicated quan
tities containing parentheses, or other signs of inclosure.
Expand:
[ (a + b) + (c + d) ] [ (a + b)  (c + d) ]
Solution :
[ (a + 6) + (c + d) ] [ (a + b)  (c + d) ] =
(a + 6) 2  (c + d)» 
a* + 2a6 + b*  (c 2  led + d 2 ) =
a* + 2ab + 6 s  c* + 2cd  d 2 .
Expand :
12. [(a + b) + (m + ri) ] [ (a + 6)  (m + n) ].
13. [ (a + 6) + 2] [ (a + 6)  2].
' (2x + J/) + 5] [ (2x + y)  5].
[ (a + 6)  2bc] [(a + b) + 26c]..
The rule may also be applied when the quantities inside
the parentheses are not of the first* degree; thus,
Expand:
[ (a 2 + 6 2 )  26c] [ (a 2 + 6 s ) + 26c]
Solution :
[ (a 2 + 6 2 )  26c] [ (a 2 + 6») + 26c ] = (a 2 + 6 2 ) 2  46V
= a* + 2a 2 & 2 + 6 4  4& 2 c*
Expand :
14.
16.
16.
17.
(a 2 + & 2 )  3] [ (a 2 + 6 2 ) + 3].
; (c + 2d)  2o6] [(c + 2d) + 2a&]
* Quantities in which the highest exponent is 2, are of the second degree.
GENERAL MATHEMATICS 127
Not only may literal expressions be multiplied, but they
may also be factored.
Problem : Factor a 2 + 3a + ab
Solution: a 2 + 3a + ab = a (a + 3 + 6)
Problem: Factor 6a 3 6 + 3ab 2 + 3ab  12a 2 6 2
Solution: 6a 3 6 + 3ab 2 + 3ab  I2a 2 b 2 =
3ab(2a 2 + b + 1  Aab).
Exercise
Factor:
1. 2a + 2b + 2c
2. ix + h{y + z)
3. xy + x 2 — Ax
4. mhi — mn 2
6. d 2 d  ab 2 + 6 3  b 2
6. a (a + 6)  b (a + 6) + 3 (a + b)
Many expressions may be resolved into their factors by
reversing the laws for multiplication stated in the earlier
part of the chapter.
Problem : Separate a 2 + 2ab + b 2 into its factors.
Solution: a 2 + 2ab + b 2 = (a + b) (a + 6)
Separate into factors:
7. a 2  2ab + b 2
8. x 2 + Axy + 4y 2
9. a 2 + 10a + 25
10. m 2  12m + 36
11. 4a 2  Aab + b 2
12. 25x 2  100z + 100
13. x 2 + 2x (a + 6) + (a + ft) 2
14. (a 2 + ft) 2 +*2 (m + n) (a 2 + 6) + (m + n) 2
15. 4 (a + c) 2 + 4 (a + c) + 1
Note: Problems 7 to 15 are really problems in square
root.
128 GENERAL MATHEMATICS
When the expression to be factored is the difference of two
squares, its factors may be obtained by reversing the rule
for finding the product of the sum and difference of two
quantities (142a). Thus,
Problem :
Factor a 2  6*
Solution :
a 2  6* = (d + 6) (a  6)
Factor:
16. x* — y*
17. a 2 — 46*
18. a 2  (6 + c)*
19. x 2 — (y  z) 2
20. (a  6) 2  (c + d) 2
21. (a 2 + 6)*  (c + d) 1
22. e 2  (/* + g 2  A 2 ) 2
The factors of a number may be readily checked by mul
tiplying the factors. The product so obtained should be the
original expression.
From the above problems it is evident that the rule for
finding the product of the sum and difference of two quan
tities (142a) may be reversed as the following rule for factor
ing:
142b. Rule: An expression composed of the difference of two
squares may be factored into the sum of the square roots of these
squares, and the difference of the square roots of these squares.
The student should f amiliarize himself with the steps in
the next example. It will be used later in the derivation
of Heron's Formula for finding the area of a triangle when
its three sides are known.
Problem: Factor: 4c 2 6 2  (c 2 + b 2  a 2 ) 2
Solution :
4c 2 6 2  (c 2 + b 2  a 2 ) 2 = [2cb  (c 2 + b 2  a 2 )] [2cb +
(c 2 + 6 2  a 2 ) ].
GENERAL MATHEMATICS 129
Removing parentheses (since the parenthesis in the first
factor is preceded by a minus sign, all the terms within it
are changed in sign when the parenthesis is removed; the
parenthesis in the second factor being preceded by a plus
sign, no change of sign is necessary) :
[2cb  c 2  ft 2 + a 2 ] [2cb + c 2 + ft 2  a 2 ].
Since changing the order of the terms does not alter the
value of an expression the quantities within the brackets
may be rearranged, thus,
[a 2  c 2 + 2cb  ¥] [6 2 + 2cb + c 2  a 2 ].
Since quantities may be put in a parenthesis preceded by a
minus sign, if they are first changed in sign (so that the two
changes of signs will cancel each other when the parenthesis
is again removed), the above expression may be restated as
[a 2  (c 2  2cb + 6 2 ) ] [ (6 2 + 2cb + c 2 )  a 2 ].
Since the quantities in the second bracket were enclosed in a
parenthesis which was not preceded by a minus sign, no
change of sign was necessary.
Since the quantities within the parenthesis are perfect
squares, they may be expressed as follows,
[a 2  (c  ft) 2 ] [(6 + 'c) 2  a 2 ].
Arranged in this form it is evident that the quantity within
each bracket is the difference of two squares, and may
accordingly be factored (142a). (It is not necessary to retain
the brackets, as the quantities within them have been fac
tored.)
{a(cfc)} { a +(c6)} {(6 + c)a} {{b+c) + a}
The parentheses may now be removed. The expression
appears, finally, as
[a  c + b} {a + c  6} {6 + c  a} {b + c + a}
130
GENERAL MATHEMATICS
X. Shop Exercise
transit
Begin the construction of the Transit, as shown by the
accompanying diagrams, which are selfexplanatory. In
stead of marking the graduations for the horizontal and
vertical circles directly on the wood, it will be found more
satisfactory to use paper protractors, which are listed at
very moderate cost in the catalogs of many dealers in engi
neering supplies. One of these will be necessary for the verti
cal circle, and two for the horizontal.
Notice that no tripod (threelegged base) is shown in
this set of drawings — since the farm level has a removable
top and the base can be used with this transit.
4 ■»
. v. [
^/Stfo/m
J—* F
6 'frotrac ttor
'ff'/aJcf
3 '/*fro tractor
^s~7k/&scopt.
'Jb"Hok
'/x m tfo/€ /br reading
arrj/e.3
*fa m MnAer
^/fvsutjhf fff
■yb'*e£3°/t
&
Zero /farA'
Winy ft*?
JLe,re/
^Pro'/'r^ao ttor 6
3/b frotroc "for
sJianc/arcj
/Lew/
Zero A f arfr
E5.fi
TRANSIT
CHAPTER XI
THE RIGHT TRIANGLE
143a. The Complement of an angle is that angle which
itiust be added to it to make their sum equal to ninety degrees
or a right angle. Hence,
143b. Rule : To find the complement of an angle subtract
it from SO degrees.
143c. Two angles whose sum is equal to a right angle are
called Complementary Angles.
143d. The complements of the same or equal angles are
equal (32b).
Exercise
Find the complements of the following angles. (Hint:
Hemember that if necessary 90° may be written as 89° 60'
or 89° 59' 60". See examples 6 and 13, Chapter II.)
l. 60°
6. 82°
11. 41° 30"
2. 45°
T. 11° 40"
12. 48° 10" 21'
3. 47°
8. 29° 34"
13. 52° 15" 28'
4. 59°
9. 32° 48"
14. 3° 0" 42'
5. 64°
10. 37° 35"
15. 72° 59" 7'
144a. If one angle of a triangle is a right angle, the tri?
angle is* called a right triangle.
The symbol for right triangle is rt. A.
144b. The side of right triangle which is opposite to the
light angle is called the hypotenuse.
144c. The sides of a right triangle adjacent to the right
angle are called the legs.
133
134
GENERAL MATHEMATICS
145a. An acute angle is an angle less than a right angle.
146b. An obtuse angle is an angle greater than a right
angle.
147a. Theorem: A triangle cannot contain more than <m
right angle.
Proof: Since the sum of the three angles of a triangle
equals two right angles (80a), it is impossible that two angles
of a triangle should be right angles, since that would leave
nothing for the third angle. Q. E. D.
147b. Theorem : A triangle cannot contain more than orue
obtuse angle.
148a. Theorem : The area of the square erected on the hypot
enuse of any right triangle is equal to the sum of the areas of
the squares erected on the legs.
K;
GENERAL MATHEMATICS 135
Given : In the right triangle PYT, FT is the hypotenuse,
nd YP and PT are the legs. YB is the square erected on
r T, and PF and YD are the squares erected on PT and YP
espectively.
To Prove : Area YB = area PF + area YD.
Proof: Draw PH _L to AB, meeting YT at G.
Part I
Draw AP and CT.
To show (a) the area of YAHG = 2 area A A YP;
(b) area CYPD  2 area A CFT; •
(c) A A YP = A C YT,
rom whence it follows that
(d) area C YPD = area AHGY (34a).
(a) Produce C Y and draw T J _L to C Y produced.
YPT is a rt. Z (given).
Also YPD is a rt. Z (121c).
.. DPT, the sum of YPD and KPT, is a st. Z (15).
.. DP T is a straight line (40).
That is, since CJ and DT are CY and DP produced,
CJ is // to DT (8g; 121c).
Now P F is ± to CJ (121c).
And TJ is J_ to C J (const.).
/. P Y is // to TJ (60a).
.'. P y = TJ (122c).
Now area C YPD = CYX YP (131).
And area CYT = C Y X TJ (133a).
Substituting for 2V its equal, FP,
Area C YT = £C Y X FP.
That is (since area C FPD =C7xyP), area C YPD =
area CYT.
(b) Produce AY", and draw PK ± to 47 produced.
By construction, Aff(? F is a rectangle (62, 121b).
.. YG and KP are both _L to A Y.
.'. YG and KP are / '/ (60a).
136
GENERAL MATHEMATICS
Also KA and PH are // (60a).
.. YG = KP (122c).
Area AHGY  AY X YG (131).
And area AFP J(i7x #P) (133a).
Substituting for XP its equal FG,
Area AYP =  (A 7 X FG).
That is, since area A HGY = AYXYG,
Area of rectangle A HG Y = 2 area of A A FP.
(c) In A A FP and C FT,
AY = YT (121c).
C F = YP (121c).
Moreover, Z C YT = 90° + Z P FT.
And Z A FP = 90° + Z P FT.
.. Z. CYT = Z AYP (34b).
A C FT = A A FP (76).
Then, by (a) and (b).
(d) Area A HG Y = area C YPD (34a).
Note: To avoid confusing the student by the great number of lines used
in the proof a separate figure is given for Part II, but after learning the
theorem the student should be able to prove all of it from one figure. Each
step in the proof of Part II corresponds to a step in Part I.
Part n
Draw PB and YF.
To show (a) area GHBT = 2 area A PTB;
(b) area PTFE = 2 area A YTF;
(c) A PTB = A YTF,
from whence it follows that
(d) area GHBT = area PTFE.
(a) Z YPT is a rt. Z (given).
Z TPE is a rt. Z (121c).
a Their sum Z FP# is a st. Z (15).
.. YPE is a st. line (40).
Again, since JT is X to C7 it is _L to DT (62).
That is, JTD is a rt. Z (12a)
GENERAL MATHEMATICS
137
and Z PTF is a rt. Z (121c).
;. Their sum, Z JTF is a st. Z (15).
.. JTF is a st. line (40).
YPE and JTF are // (8g, 121c)
and CJ and D3P are // (Part I). .
.. YJ  FT (122c).
Now area PTFE ~ PTX TF (131).
And area YTF = § (YJ X TF) = * (PT X T^). (133a.)
That is, since area PTFE  PT X TF, area square
/TFE = 2 area A FTF.
(b) Draw PL ± to BT produced.
GHBTisa rectangle (62, 121b).
Then GIT also is _L to BL.
138 GENERAL MATHEMATICS
.. GT is // to PL (60a).
Furthermore PH and BL are both J_ to AB.
:. PH is // to BL (60a).
.. GT  PL (122c).
Now the area GHBT. = GT X TB (131).
And area PT£  £ (PL X TB) (133a).
Substituting for PL its equal GT,
Area PTB =%(GTX TB).
That is, since area GHBT =GTX TB,
Area rect. GHBT = 2 area A PTB.
(c) Now in A PTB and YTF
PT = !TP
(121c) 1
YT = TB
(121c) I
Z PTB = 90° + Z PT Y.
Z FTP = 90° + Z PZ 7 7.
z ptb = z rrp
(34b)
A PTB = A FTP
(76)
Then, by (a) and (b),
(d) Axe&GHBT = area PTFE.
Part m
I. Area YABT = area AHGY+ axenGHBT (32a).
But area A HG Y  area C FPD (Part I).
And area GHBT = area PTF# (Part II).
Substituting these values in the above equation (I),
Area YABT  area C YPD + area PTFE. Q. E. D.
The fact that the square of the hypotenuse of a right
triangle equals the sum of the squares of the two legs was
known thousands of years ago in Egypt, where it was used in
laying out the corners of buildings. The modern carpenter
does practically the same thing when he lays out the sills of
a house (the beams which rest upon the masonry founda
tion), by measuring along an end beam 6 feet from one cor
GENERAL MATHEMATICS 139
ner, 8 feet from the same corner along the side beam and
then laying a strip 10 feet long, so that one end coincides
with the 6 foot mark, while he moves the other beam until
the 8 foot mark coincides with the end of the 10 foot strip
— thereby laying out the "6, 8, 10 triangle," and so making
the angle between the beams a right angle.
Although this property of the right triangle has been
known from remote antiquity, no general proof of it appears
to have been given until the following theorem (148a) was
demonstrated by Pythagoras, in whose honor it was named.
Pythagoras was a famous philosopher and mathematician
who was born in the Grecian island of Samos about 582 B.C.
and died in Magna Grsecia about 500 B.C.
Pythagoras studied at Miletus under Thales. A generally
credited tradition states that he spent several years in study
in Egypt, and journeyed to Babylon and even to India. (The
nature of his work in mathematics corroborates this.) Pytha
goras thus learned all that was then known of mathematics.
In later life he founded a famous school at Crotona in South
ern Italy. He established a secret society, the members of
which were known as the Pythagoreans. The Pythagoreans
did much for the advancement of knowledge.
As the student knows from Arithmetic, the sign \T~
before a quantity means that the square root of the quantity
is to be found. This sign is read "the square root of."
Thus the equation, _____
a = V c 2  b 2
is read, "a equals the square root of c square minus b square."
The equation,
3 = ^9
means that 3 is one of two equal factors whose product is 9.
Solve :
(a) (b) ^ _(d)_ (e) (f) (g)
1. V 25 V 64 V 100 V 144 V~169 VloO VI96
140 GENERAL MATHEMATICS
2. Carry out to two decimal places by the method learned
in arithmetic:
(a) (b)
V740 V1982
For convenience in computation the property of the right
triangle proved in the Pythagorean theorem may be stated
as follows:
148b. The hypotenuse of a right triangle equals the square
root of the sum of the squares of the two legs (134a).
Letting c represent the hypotenuse, a one leg, and 6 the
other leg of a right triangle, the Pythagorean theorem might
also be expressed as
I. c* = a* + 6*
Whence, c = V a * + 6* (134b)
From (I) b y transpo sition, c* — a 1 = 6 s ,
Whence, < <*  a* = b (134b)
Or, 6 = < <*  a*
By a similar derivation, a = V c 2 — b*
That is,
148c. Rule : Either leg of a right triangle equals the square
root of the difference of the squares of the hypotenuse and the
other leg.
Exercise
1. One leg of a triangle is 10 feet long and the other is
18 feet long. Find the length of the hypotenuse.
2. The hypotenuse of a right triangle is 20 feet long and
the base is 12 feet long. Find the altitude. (Remember
that the altitude of a right triangle coincides with one leg.)
3. How long a ladder is required to reach to the top of a
wall 22 feet high if the lower end of the ladder is to be placed
9 feet from the wall?
GENERAL MATHEMATICS 141
4. A flag pole is to be secured by three cables passed
through holes in an iron collar 24 feet from the ground.
Hie lower ends of these cables are passed through rings
Fastened in concrete bases at the vertices of an (imag
inary) equilateral triangle. If each of these rings is 10
feet from the flag pole, how many feet of cable must be
ordered, allowing 2 feet on each cable for fastening and
slack of cable?
Note : The eaves of a house is the line where the roof
joins the side walls.
The rafters are the sloping timbers to which the roof
soards axe nailed.
The timber which runs along the highest part of the
xamework of the roof, to which on either side are nailed
ihe upper ends of the rafters, is called the ridge pole.
5. The height of a house from the foundation to the eaves
s 18' 6" and the height from the ridge pole to the foundation
s 26' 10". The width of the house is 21'. Find the length
>f the rafter. Adding one foot to each rafter for waste in
cutting find how many linear feet must be ordered for 13
Dairs of rafters.
6. It is desired to make a tent by putting canvas over a
evel ridge pole 10 feet long and 7 feet above the ground,
;he tent to be 8 feet wide at the bottom, the ends of the
;ent left open. How many square feet of canvas must be
>rdered for the tent?
As the student already knows, the symbol V~~ may be
ipplied to quantities which are not perfect squares — as, V5.
148d. The expression for finding the square root of a quan
tity which is not a perfect square is called a quadratic surd.
Because there is no number, which multiplied by itself will
;ive 5 — in other words, since 5 has no exact square root, it
ioes not follow that V~5 has no definite meaning.
142 GENERAL MATHEMATICS
Thus,
Problem: Draw a line V5 inches in length.
Solution: Draw a rt. A, one leg of which shall be 1 inch
in length, the other 2 inches. Then (by 148b),
Hypotenuse = V 1* + 2 *
 Vl + 4
= V5
The length of the hypotenuse, then, is V5 inches,
and by setting the points of the dividers at the two extremi
ties of this hypotenuse, its length, V5, may be transferred
to any diagram desired. It is evident from this that V5
has a definite meaninglmd is capable of exact representation.
Problem : Draw a line V3 inches long.
Solution: Draw a rt. Z. On one of the sides lay off a
point 1 inch from the vertex, and with this point as a center
and a radius of 2 inches, describe an arc intersecting the other
leg. This point of intersection determines the length of the
second leg. Draw the hypotenuse. By construction, one
leg of the triangle is 1 inch in length and the hypotenuse is
2 inches.
/. the second leg = V4  1 = V~3
7. Draw lines having the following lengths in inches:
(a) V 7 Hint: V4*  3* (b) V~io
(c) VI5 (d) VI7
(e) VI9 Hint: V 10*  9* (f) V~21
(g) ^34 (h) VTT
Note: In this chapter a quadratic surd will be called "a surd."
Many surds may be easily represented in two ways: thus,
V~13 = V 2 2 + 3* or V 7 2  6*
Problem : Draw a line 3 V~5 inches long.
GENERAL MATHEMATICS 143
Solution : Draw an indefinite straight line. Construct a
line V~5 inches long. Take off this length between the points
of the dividers, and lay it off 3 times along the indefinite
straight line: obviously the length so determined will be
3 VI) inches.
8. Draw a line:
(a) (b) (c)
2 V io in. 3 VT7in. 4 V~7in.
It is often possible t o sim plify surds: thus,
Problem: Simplify V~2a
Solution: V 20 = V 4X 5 = 2< 5.
Proof : Construct a line *v20 inches long by drawing one
leg. of art. A 2 inches long and another leg 4 inches long.
The hypotenuse will be V~20 inches in length. Construct a
line V~5 inches long,' and taking this length b etwe en the
points of the dividers lay it off along the line V 20 inches
long. It will thus be found that the V~5 will go just twice
into V~20. The following problems may be proved in the
same way.
The square root sign (as well as the cube root sign, etc.)
is often called the radical sign.
148e. Rule: Surds may be reduced if it is possible to
separate the quantity under the radical sign into factors, one
or more of which are perfect squares. The square root of these
perfect squares may be extracted and their product written out
side the radical sign as the coefficient of the surd.
Problem: S implify V I80a 3 6
Solution: V I80a 3 6 = V9 X 4X5Xa 2 XaX& =
3 X 2a Vftoft = 6a V 5<z6
9. Simplify:
W 0>) (c) (d) (e)
V 75 V^3 V 4a 2 z V ( a + 6) 3 V 2 (a 2  2a6 + 6 2 )
144 GENERAL MATHEMATICS
(0 (g) GO (0
V~27 V~48 Vl62 V 4 (a + 6)* (a  6)V
Problem: Simplify
/r
Solution:
i/ 8 *** j/46»X2 26 ,—
3
(a) (b) (c)
10.
. / 36c*d* , / 75x» , /3 , , w
y — y — fi (fl+i)
(a) (b) (c)
11.
A /49c* . /816V , / 10 (m + n ) ]
y "9" y ~6T" y §6
(a) _00 (c)
i4 f/f 1 /^
In case the denominator of the fraction under the radical
is not a perfect square it may be made so by multiplying
both terms of the fraction by the same number: thus,
Problem: Simplify
>,y Vi
Solution:
\T\\r\'\/\™=\\r*
GENERAL MATHEMATICS
145
(a)
18. Simplify:
ft
(b)
ft
(c)
14.
V
(a)
8c
(b)
3
(a)
16.
A
+ b
(b)
/
(c)
5
(c)
1^
(a  &)»
149. The projection of one line on another is that length
on the second (produced if necessary) included between per
pendiculars let fall upon it from the ends of the first.
Thus, the projection of AB on GL is the length of PN,
the segment of GL included between perpendiculars let fall
upon GL from A and from B.
When one end of a line meets the line on which it is to be
projected, the same definition applies if we consider one of the
perpendiculars as of length zero.
Thus, the projection of A B on GL is the length AC, the
perpendicular from A to GL being of zero length.
146
GENERAL MATHEMATICS
Exercise
l* Draw a triangle with sides 5, 434, and 3 inches long,
respectively, and find the projection of each side upon the
other (produced if necessary) by drawing the perpendiculars
and me9suring the projection with the ruler.
2. In the triangle ABC, find the length of the projection
of side AC on side AB, of side AB on side AC, given that:
AB = 16, BF = 8, AC = 9J, DC = 5 (148c).
r^
tl
3. In the triangle MNP, find the length of the projection
of the side MP on side MN, and side MP on side NP, given
that MN = 14, QP = 4, M P = 7, M J =6.
150a. An Obliqt*? Triangle is one that does not contain a
right angle.
150b. Theorem: In any^ligue triangle the square of the
side opposite an acute, angle is equal to the sum of the squares
of the two sides adjacent to the acute angle minus twice the
product of either one of these two sides by the projection of the
other upon that side.
GENERAL MATHEMATICS
147
Case I
(When the projection of the side opposite the acute angle
sJls upon the base.)
Given: In A ABC, L A is acute and a is side opposite
^ A ; b and c are the sides opposite L B and Z C, respectively;
<4P the projection of c on b.
Part I
To Prove: a* = 6* + c 2  2AP(&)
Note: The bar is used over AP to mean "the line AP."
Proof: PC =b  AP.
Squaring both sides of this equation:
PC* = &*j 2 AP (&) + IP*
Adding BP* to both sides of this equation:
(I) BP* + PC* =¥  2AP (&) + AP* + BP*
But Pfl 2 _+ PC 2 = a 2 (148a).
AP* + BP* = & (148a).
Substituting these in (I)
a* = 6 2  2lP (6) + c 2
That is,
o* = &* + c*  2AP (6).
148
GENERAL MATHEMATICS
Part II
Note: Since the theorem makes the statement concerning "either one of
the two sides adjacent to the acute angle/' it is necessary to prove that it
holds true for both.
Given: AF is the projection of 6 on c in above triangle.
To Prove: a 2 =_6 2 + c 2  2AF (c)
Proof: BF = AF  c
Squaring both sides of this equation :
BF 2 AF 2  2AF (c) + c 2
Adding FC 2 to both sides of this equation:
(II) JIF 2 + ¥& =AF 2 + FC* 2AF (c) + c*
But BF 2 + FC 2 = a 2 (148a)
and AF 2 + FC 2 = 6*
Substituting in (II)
a 2 = 6 2  2AF (c) + c 2
That is, _
a 2 = 6 2 + c 2  2AF (c)
Case II
When the projection of the side opposite the acute angle
falls upon the base produced.
4,
Part I
Given: In A ABC, Z C is acu^e, Z B is obtuse; sidec
is opposite Z C, side 6 is opposite Z B and side a is opposite
Z A ; CP is the projection of 6 on a.
GENERAL MATHEMATICS 149
To Prove :
c 2 = a 2 + b 2  2a (PC)
Proof :
PB = PC  a
Squaring both sides:
PB 2 = PC 2 2a {PC) + a*
Adding AP to both sides of this equation
AP 2 + PB 2 = AP 2 + PC 2  2 (a) PC + a 2
The remainder of this proof is left to the student.
BD is the perpendicular needed in Part II.
Exercise
In the following exercises, let the student draw the figures
carefully to scale.
1. In A ABC, find the length of a (opposite the acute
Z A), given c (or AB) = 12; 6 (or AC) = 8' and AD (the
projection of c on b) = 10 1/2'.
Check. The same answer should be obtained by using
the projection of b on c. AE (the projection of b on c) = 7.
150c. Theorem: In any oblique triangle, the square of the
side opposite an obtuse angle is equal to the sum of the squares
of the two sides including the acute angle, plus twice the prod
uct of either one of these including sides by the projection of the
other upon that side.
Given: In A ABC, Z A is obtuse and a is side opposite
Z A; b and c are sides opposite A C and B respectively: AD
the projection of con b.
To Prove: a 2 = b 2 + c 2 + 2b (AD)
Proof: CD =b + AD.
Squaring both sides of this equation, etc.
The construction of the figure according to the above direc
tions and the remainder of the proof of this theorem (which is
similar to that of the preceding theorem) , is left for the student.
150 GENERAL MATHEMATICS
2. In A MNQ find m, given that q = 32, n = 42, and MS, In
(the projection of n on q) = 33. ISi
Check: Practically the same answer should be obtained
by using MR (the projection of q on n) = 25.14.
XIa. Shop Exercise
stairs
Definitions : The treads are the level boards on which the
foot rests when ascending the stairs. The risers are the
boards perpendicular to the treads. The stringers are the
sloping supports to which the treads and risers are nailed.
Procedure : Plane one edge of a piece of white pine and
lay off 15 in. along the edge. With one extremity of this
line as a center and a radius of 9 in. describe an arc, and
with the other extremity as a center and a radius of 12 in.
describe a second arc intersecting the first. From this point
of intersection draw lines to the extremity of the 15in. line,
thereby forming a right triangle. (Why?) Cut out the
board along this line.
Make a strip of wood an inch wider than the thickness of
the triangle and 17 in. long, and nail it along the 15inch edge
of the triangle to serve as a guide.
/ "*t Tn T^r t*
T
End  *^ C I CQ + n Q t fed on
Lay this triangle on the board which is to be the stringer
(which should be at least 9 in. wide) so that the strip is flush
with the edge, and mark along the other two sides. These
GENERAL MATHEMATICS 151
pencil marks, with the edge of the board, form a right triangle.
Then slide the pattern along and mark another triangle, just
bouching the first. Continue this process until six of these
triangles have been laid off.
The base of the first triangle should be produced to meet
the other side of the stringer to form the bottom line. The
altitude of the last triangle should be produced to meet the
edge of the stringer and form the top line.
Evidently when the stringers are in place, and the lower
treadboard, 1 in. thick, is nailed on, the height of its top
above the floor will be 10 in. This thickness is subtracted
from the height of the second riser, but when the second
riser is put in place this adds the same amount as was sub
tracted, so the second riser is still 9 in. in height. This will
be true of all the remaining risers; only the first will be 10 in.
in height. To remedy this, lay off a line parallel to the
bottom line and 1 in. (the thickness of the treadboard)
above it. Cut on this line.
When the first riser board is nailed on, the width of the
first tread is increased 1 in. When the second riserboard
is nailed on, the same amount is subtracted, so the width of
the first tread remains the same. This is true of all the treads
up to the last. When the last riser is nailed on, however,
the width of the last tread is thereby increased to 10 in. and
there is no other riser to be subtracted from it. To remedy
this, draw a line on the stringer parallel to the top qut, and
1 in. (the thickness of the riser) within it. Cut along this line.
Cut out the triangular pieces as marked, and the stringer
is complete. The remaining stringer may be marked with
this as a model. See Xlla.
Xlb. Instead of making the above exercise full size,
a model may be nuade by dividing all the above dimensions
by the same number: for example, instead of laying out a
triangle whose sides are 9, 12, and 15 inches, one may be
laid out whose sides are 3, 4, and 5 inches long.
CHAPTER XII
LITERAL FRACTIONS: HERON'S FORMULA
151. Literal Expressions may not only take the form of
Integers, but of Fractions, as well.
The terms learned in Arithmetic (Numerator, Denomina
tor, Common Denominator, etc.) apply equally as well to
literal fractions, and all the operations with fractions
learned in Arithmetic may equally as well be performed
with literal fractions.
Exercise
Reduce the following fractions to lowest terms:
l. 36sV3 2
24x 2 i/3 3
Solution :
36s 4 i/ 3 s 2 _ 332 2x 2 x 2 yy 2 z 2 3x*y 2
24x 2 yz* 3 • 2 • 2 • 2 • x 2 • y • z 2  z = 2z
Note : A dot placed up midway of the letter so that it does not look like a
decimal point indicates multiplication.
Factors common to both terms are cancelled as in Arith
metic. In the solution, instead of writing the x 4 as x • x • x • x • ,
it is better to write the highest power of x to be found in
both terms as one factor, that is, to write the x term in the
numerator as x 2 • x 2 , for a glance at the denominator shows
that there is an x 2 there to cancel it.
Likewise the y term of the numerator is separated into
y • y 2 , since there is a y in the denominator, while the z
152
GENERAL MATHEMATICS 153
term is written as z 2 , since there are sufficient z factors in
the denominator to canpel it.
2. 12a 3 fr 2 3. 7x*y* 4. l&nthip*
36a6 lAxy 2 z
6. 120a 3 b 2 c 6. 50x*yz*
20win 2 p
7.
39ax*
13ox 2
10.
54o 2 &
75ab 3 c 2 15x z y 2 z*
8. 88x*yW 9. 63^6^
66r4/s 2 84a 4 6 4 c 2 99a 2 6 3 c
Problem: Reduce a 2 + 2ab
ab
Solution:
a 2 + 2ab _ a (a + 2b ) = a + 26
a& a • 6 6
11. b 2 + 4ab 12. 3a: 2 + Qa x
b 2 12ax
Hint: Do not make the mistake of considering ax as
factor of both numerator and denominator in Example 12.
Problem: Reduce a 2  2ab + b 2
a 2  b 2
Solution :
a 2  26b + b 2 (a  b) (a  b) a  b
a 2  b 2 (a  6) (a + 6) ~~ a + 6
13. x 2 + 2s2/ + y 2 14. x 2 + Axy + 4y*
x 2 — y 2 . a: 2 — 4j/ 2
16. q 2 rc 2 + 2a fogy + by
ax + by
154 GENERAL MATHEMATICS
Least Common Multiple
152a. It was learned in Arithmetic that in order to add
fractions it is necessary to change them to a Common
Denominator and in order to find the Least Common
Denominator of several fractions it is necessary to find
the Lowest Common Multiple, and so was learned a method
for finding the Least Common Multiple of any given number.
For the same reason, it is necessary to find the Least Com
mon Multiple of given literal expressions. The method
for finding the Least Common Multiple of given literal ex
pressions is fundamentally the same as that in Arithmetic,
but is shortened somewhat by the aid of exponents, which
save the work of repeating a factor. Thus, in Arithmetic,
the factors of 8 would be written 2x2x2, which by the
aid of exponents may be written 2 3 .
Problem: Find the Least Common Multiple of:
15a 2 6c, 20a 3 6 2 c, 30a be 2
Solution :
15a 2 &c = 3'5aa b c
20a 3 & 2 c = 225aaa &• bc
30a&c 2 = 352a6cc
As in Arithmetic, take each factor the greatest number of
times it occurs in any one of the three quantities to be
factored.
3225a a <*•&•&• cc= 60 a 3 6 2 c 2 , L. C. M.
By aid of exponents, we may avoid repeating factors—
that is, separating a 3 into a' a' a', b 2 into b' 6, etc., by aid of
the following rule:
152b. To find the Least Common Multiple of given literal
quantities find the Least Common Multiple of their numer
ical coefficients and then take each literal factor with the
highest exponent that it is contained in any of the given literal
quantities.
GENERAL MATHEMATICS 155
Thus, in the above example, we find the Least Common
Multiple of 15, 20, and 30 (which is 60) then write a with 3
as an exponent, since a 3 is the highest power of a found
in the given quantities (that is, the a factor which has the
largest exponent), then write 6 with 2 as an exponent (since
b 2 is highest power of b found in the given quantities) and
finally write c with 2 as an exponent (since c 2 is the highest
power of c found in the given quantities) making the final
answer 60a 3 6 2 c 2 . The process is the same when the quan
tities contain Binomial or Polynomial Factors.
Exekcisb
Find the Least Common Multiple of the following :
1. 7x 2 y, 3x*y 2 x } 2xy 2 z 2
2. 4a 2 be, 8a 3 6 2 c, 12a 6 2 c 3 , 16a 8
3. 6m 3 , 15n 3 , 18p 3
 4. 24a 3 #, 40ah, 50a 2 h
6. 32a6c, 16a 2 V8& 2 c*
6. 3m 3 , 9n, 18p 2 , 24mrc 2 p
7. 5x 3 y z z 2 , 15zy% 25x 2 y 2 z
8. 6 (a + 6), 4 (a  6) 2 , 2 (a 2  b 2 )
Solution: The L. C. M. of the numerical quantities is 12.
We find that (a + b) occurs only once in any quantity, and
(a — b) occurs twice as a factor: The L. C. M. is therefore
12 (a + b) (a  6) 2 . Note that the exponent in this ex
pression applies only to the factor which it follows — that
is, to (a — b) and not to (a + b).
9. 3 (x  y), 2 {x 2  j/ 2 ), 5 (x + y)
10. 4 (a  6), 2 (a + 6), 3 (a 2 + 2a& + b 2 ) .
11. 5 (x  2y), 10 (x + 2y), 15 (x 2  4y 2 )
12. 2a, 3 (a  26), 2 (a + 26)
18. a 2  46 2 , 3 (n  6), 5 (n 2  6 2 ), (n + 6)
14. 2 (6 2 + 26c + c 2 ), 3 (6 2  26c + c 2 ), 4 (6 2  c 2 )
156 GENERAL MATHEMATICS
In Arithmetic it was learned that before fractions of dif
ferent denomination can be added or subtracted they must  i
be changed to the same denominator. For example:
Add: 2 3 5 1
3 + 4 + 6 + 2
Solution :
2 8 3 9
 can be changed to — >  to — > etc., without change of value,
since the value of a fraction is not altered if both numer
ator and denominator are multiplied by the same quantity
8 24 2
that is,  = — = 
Hence, ^_,_9,ip,^ = 33 = ll =
12 + 12 + 12 + 12 " 12 ~~ 4 4
This principle apphes equally well to literal fractions.
Combine: 2 3 1 1
— r — ; + 7T —
a a* 2 4a 2
Solution :
By inspection we find that 4a 2 is the L. C. D. and that a
must be multiplied by 4a to make it equal 4a 2 . We there
2
fore multiply both terms of  by 4a. Thus:
a
2 _ 8a
a " 4a 2
3
We find also that a 2 , the denominator of — , the next frac
a 2
tion, should be multiplied by 4 to make it equal to 4a 2 . We
3
therefore multiply both terms of — by 4. Thus:
1 = 11
a 2 4a 2
GENERAL MATHEMATICS 157
We find, also, that 2, the denominator of the next fraction,
should be multiplied by 2a 2 .
We therefore multiply both terms of — by 2a 2 . Thus:
\_ _ 2a 2
2 ~ 4a 2
The last fraction — > does not need change of form. The
4o 2
example now appears as follows:
23JL_J_ = 8ol22o*_J^
a a 2 2 4a 8  4a* 4a 2 4a 2 4a*
8a + 12 + 2a 2  1 2a 2 + 8a + 11
~ 4a 2 ~~ 4a 2
Problem :
Combine: 4x + 6 3x  5
" +
Solution:
8 ' 6
4x + 6 ' 3x 5
8 + 6
= 3 (4a; + 6) 4 (3x  5)
24 + 24
12x + 18 + 12x  20 _ 24x^2
24 24
2 ( 12x  1) 12x 1
2 12 ~ 12
Exercise
Combine:
2_ _3_ _ 26 _3_
*' 66 + 26* " 3 + 26
158 GENERAL MATHEMATICS
J_ W h _ 2h
*' 4 4c* c* 2c*
3 2 3
3. — + .
x* xy xy*
4 * 6* + 126* ~ ab + 3a*
Z_ _ 26 5a Sab
*' ab ~ 5a + 26 ~ To
. 3x* , x* 1 2,3
6. — 1
4j/* 2y* x*j/ s x y
3 3 6* a*
7 ' 2a6 + 7a*6 14a + 6
3 4m m n
8. : 1 + r. +
mtnp np*m n*p* p*ro m*np
3 7,4,5
9. —
x x* 2x 2x* '
_2_ 3x _ 3x _2
10# xy* 2y l 2 xy
_4_ _ 36 6^
6c 4c* c*
2x + 9 , 6x  5
*' 5" + ^5
2x  3 3x + 8 9x  4
"* 4 " 6 + 3
3x  2 7x  8 5x + 2
"• 4 + 6 ~ 15
2 (6x + 5) 3 (x + 6) 4 (3x  4)
16  3 + 9 6
GENERAL MATHEMATICS 159
2x + 1 3z , 4 , X
16 ^f5+10 + 2
3a; +7 2 (g  1) 8_
* 17 ' 4 + 22 "11
In Arithmetic we learned that in writing a mixed number,
that is, an integer plus a fraction, it is unnecessary to write
the plus sign between the whole number and the fraction —
that is, 5 + $ may be written as 5f .
In literal expressions, however, a whole number, which
is to be combined with a fraction, must be connected with it by
a plus or minus sign — since to omit it would mean that the
fraction was to be multiplied by the whole number.
In Arithmetic we learned that an integer (whole number)
may be expressed as a fraction with a given denominator,
by multiplying the integer by the denominator, and writing
the product over the denominator.
Problem : Express 5 as a fraction whose denominator is 4.
Solution: 5 = —
4
Proof: Dividing both denominator and numerator by 4
5
We obtain  or 5.
The same method is used in dealing with literal expressions
Thus:
Problem : Change a to a fraction whose denominator is 6.
Solution: a= —
b
Proof : Cancelling 6 in both terms of the last fraction, we
have a.
Hence,
153. Rule : Whole numbers and fractions may be combined
by changing the whole numbers to fractions whose denominator
is the common denominator , and then combining as when all
the quantities were fractions.
160 GENERAL MATHEMATICS
Combine:
18.7 + 
a 2 3 ■ 2
20. 4 t + 
c b a
c* +a*
21. 6» rT
4c*
4x
Problem: 3 +
x 1
Solution : Since the bar ( — ) of a fraction serves as a
parenthesis by causing the numerator to be treated as one
quantity, and the denominator to be treated as one quan
tity, we may consider a parenthesis to be placed around
(x  1) and then multiply 3 by it, putting the product
over the denominator. Thus:
4a; _ 3 (x  1) 4x _ 3x3+4x _ 7s 3
3 f (x1) ~ (x 1) (x1) " x1 ~xl
x  2
22. 3 
28. 5 +
X1
3x + 1
x 2
24. Zy + 4 +
25. 2x  7 
2
5y + 1
5x
x +2
3x  1
26. 2x + 1 + _ „
GENERAL MATHEMATICS 161
Note : The following will occur in the derivation of the
formula for the area of a triangle called the Heronian
Formula :
Problem :
Combine :
Solution :
(6 2 + c 2  a 2 ) 2
4c
2
_ ^6M_c2_ JL a^ _ 4b 2 c 2 _ (b 2 + c 2  a 2 )* _
~ 4c 2 ~ 4c 2 ~ 4c 2 ""
4b 2 c 2  (b 2 + c 2  a 2 ) 2
4c ?
Let the student express the numerator as the product of
its prime factors. This will necessitate factoring twice.
See Chap. X.
164. Rule: Fractions occurring in equations may be elim
inated by multiplying each term in the equation by the com
mon denominator of the fractions.
Thus:
Problem: Solve for x:
3s , „ 5 x
2 +6 = 3"4
Solution: The least common denominator is 12. Multiply
each term in the equation by 12 (101).
123a; a 10 125 12a;
__ +6 . 12 =  r
Reducing fractions to lowest terms:
63a; + 72 = 45  3x
18x + 72 = 20  3a;
21x =  52 (99a).
Dividing both sides by the coefficient of x,
52
x =
21
162 GENERAL MATHEMATICS
EXEBCISB
Solve far x:
(a) (b)
21z 7
1. ==r = 42 =28
5 x
2x 1 _ 3x
** 5 + 5~ 10 " 3
4. ^^ + 5x  31 (153).
8. Solve for a: U ~ 2 +2=^+3
6. Solve for a; ^ + * + 3 = 2a  1
155. Heron's Formula : To compute the altitudes of a tri
angle in terms of its sides: (See Chap. X.)
In the A ABC, at least one of the angles is acute. Sup
pose A is acute — draw BD or A, J. to AC.
(I) In the ABDA, ft 2 = c*  AD? (148a)
In the A CAB, a 2 = c 2 + 6 2  26 X AD (150b)
GENERAL MATHEMATICS 163
Transposing a* and 2b X AD,
26 X AD= c* + b* a*.
Dividing both sides of this equation by 26,
An c* + 6*  q»
AD 26
Substituting in (I),
,, _ , [c* + 6*  a*]* _ 4c*6» [c*+ 6»a»]»
h ~° ~ 46* " 46'
[2c 6  (c* + 6*  a*)] [2cb + (c* + 6*  o*)]
46*
Dropping parentheses,
. . (2c6  c*  6* + a*) (2c6 + c* + 6*  a*)
*' 46*
Changing the signs of certain terms and then enclosing
them in a parenthesis preceded by a minus sign:
r. [a*  (c*  2c6 + &*)][ (c* + 2c6 + 6*)  a*]
h 46*
[a*  (c  6)*] [ (c + 6)*  a*]
46*
[a
7< c
 6) ] [a + (c •
 6) ] [ (c + 6) 
a] [(c
+ 6) + a]
/i* =
[a
 c + 6] [o +
46*
c  6] [c + 6 
a] [c
+ b+a]
= 1/
46*
■ h
f [a c + 6] [a
+ c  b] [c + 6
a] [c
+ b + a]
V 46^
164 GENERAL MATHEMATICS
For brevity let a + 6 + c = 2s
From a + b + c = 2s, subtract 2c and we have
a + 6 + c = 2s
2c = 2c
a+bc = 2s2c
Factoring:
a + 6  c = 2 (s  c)
From a + 6 + c = 2s, subtract 26 and we have
a + b + c = 2s
2b = 2b
a  b + c =2526 = 2 (s  6)
From a + 6 + c = 2s, subtract 2a and we have
a+ 6 + c = 2s
2a = 2a
a+6 + c = 2s  2a = 2 (s  a)
Substituting in the above value of h
h _ 4 /[2 (g  c) ] [2 (s  6) ] [2 («  q) ] [2s]
*y 46^ : —
16 (s  c) (s  6) (s  a) s
46 2
4 (s  c) (s  6) (s  a) s
" ~~~6^
2 j
h = r /(s  c) (s  6) (s  a) s
The area of A ABC = \b X h (133a).
Substituting the above value of h
Area ABC =  X 6 x ^ j/(s  c) (s  6) (s  a) s
i
GENERAL MATHEMATICS 165
Cancelling:
Area ABC = a/(s  c) (s  b) (s  a) s
Rearranging terms:
Area ABC = a/s (s  a) (s  b) (s  c)
Xlla. Shop Exercise
ERECTION OF STAIRS
Draw a pencil line on the floor making a right angle with
the wall, at that place on the floor where it is desired to
locate the stringer for the stairs. At the point of intersection
of this line with the wall, draw a pencil line on the wall,
making a right angle with the floor. Starting from the inter
section of these two lines measure off on the line on the floor
a distance equal to five times the tread of the stringer and
mark this point. Starting from the intersection of these two
lines, measure up the line on the wall a distance equal to
five times the height of the risers and mark this point.
At the distance desired for the width of the stairs lay off
similar lines and mark them in the same manner. The
stringers may now be erected, keeping in mind the allowance
of 1 in. which was made for the tread and the riser. The
riserboards and treadboards may now be nailed in place.
The treadboards should overlap the riserboards § in. in
front, and should overlap the stringers 1 in. on the sides.
Xllb. Field Exercise
AREA OF FIELD BY HERON'S FORMULA
Equipment: Leveling instrument, ranging pole, stakes,
measuring tape.
Procedure: The field should be divided into triangles,
which should be as nearly equilateral as practicable. Hence,
166
GENERAL MATHEMATICS
having measured one end of the field, set up the instrument
in one corner and lay out a line intersecting the opposite side
at about the same distance from the first corner of the field
as the length of the side first mentioned. For example,
having measured AB, lay off BC, measure from stake A to
stake C, lay off CD, DE, EF, and measure them; also
measure BC, BD, DF, CE, EM and FM . Compute areas of
triangles in square chains by Heron's Formula, find their
sum and change to acres.
Make a drawing of the field by adopting some convenient
scale, such as 50 links = 1", or 20 links = 1*, or whatever
size will best fit the paper. Lay off the length AS to scale.
Set the dividers with the length BC to scale and, with B as
a center, describe an arc; set the dividers with the length
AC to scale and, with A as a center, describe an arc inter
secting the first arc at C. Draw in the triangle lightly. In
like manner the other triangles may be drawn. The outline
of the field may then be drawn in solid lines — the other
lines being dotted.
If desired various topographical details occurring in the
field may be located by noting their points of intersection
with the various lines thus laid out. Thus, by measuring
such lines as GD, DH, DI, FJ and EK, a brook may be
located. A rock as at L, may be located accurately by
laying off and measuring the extra lines AL, BL } and LC.
CHAPTER XIII
SIMILARITY
In general, similar figures are of the same shape but not
of the same size. For example, a circle 1 in. in radius and
one 10 in. in radius are similar. Again, a rectangle 2 in.
long and 1 in. wide is similar to a rectangle 2 ft. long and
1 ft. wide.
It is evident that for two triangles to have the same shape
it is necessary that the angles of one triangle be equal
respectively to the angles of the other.
156. . In describing similar triangles, "corresponding sides"
has the same significance as in dealing with equal triangles —
that is, in two similar triangles, corresponding sides are
sides opposite equal angles.
In the example cited above, of two similar rectangles, the
altitude of the smaller is contained as many times in the
larger as the base of the smaller is contained in the base, of
the larger — that is, the ratio of the altitudes is the same as
the ratio of the bases.
157. The ratio of one quantity to a second is the number
of times the first contains the second.
Thus, the ratio of a yard to a foot is 3; the ratio of 10
to 5 is 2.
Obviously, a ratio is an abstract number. Thus, the ratio
of a yard to a foot is simply 3, not 3 inches, or other units.
158. Two similar triangles are triangles the angles of
which are equal each to each, and the corresponding sides of
which are in the same ratio.
167
168 GENERAL MATHEMATICS
159a. Theorem: Two mutually equiangular triangles are
similar.
Given: In the triangles ABC and A'B'C, Z A = Z i',
Z £ = Z JB' and Z C = Z C".
To Prove: ABC and A 'B'C' are similar.
Note: Since the 2* of the A are equal, each to each, it
is only necessary to show that the corresponding sides are
in the same ratio; that is, to show that
AB = jiC BC
A'B' A'C' ~B'C
Proof: Place A'B'C on triangle ABC so that Z A' shall
coincide with its equal, Z A, the vertex A' falling on A, the
side A' B 1 falling along AB and side A'C along AC; that is,
A'B 1 taking the position AB 1 and A'C 1 taking the position
AC.
Now Z AB'C = Z AJ5C (given).
/. B'C is // to BC (67b).
Take some unit of length small enough to be contained
without remainder a certain number of times in AB' and a
certain number of times in B'B. Let this unit be contained
m times in AB 1 ; that is, divide AB 1 into m equal parts.
At the points of division construct //s to B'C'. Then AC
will also be divided into m equal parts (123). Let the
same unit of length be contained n times in B'B — that is,
divide B'B into n equal parts. At each of these points of
division construct //s to BC.
These lines are// to B'C also (61b).
CC is divided into n equal parts (123).
GENERAL MATHEMATICS 169
That is, AB is divided into m + n equal parts and AC
is divided into m + n equal parts.
AB' m
AB m + n
AC = m
AC m + n
• m: = ^: (31)
" AB AC {61)
Or since AB' and AC" are simply other positions of A'B'
and A'C,
A'B' AJC_
W AB AC
That is, the sides including the angle A are in the same
ratio.
In like manner, placing the Z C so that it coincides with
Z C, it may be proved that A'B' is// to AB, then by
taking a common unit of length as before and drawing //s
it may be proved that
A'C B^C
AC BC
Then by I,
A'B' = WC_ = A'C
AB BC " AC
Thensince Z A = Z A' y Z £ = B' y Z C = ZC (given)
A'JB'C" is similar to ABC (158). Q. #. D.
159b. Theorem : Two right triangles are similar if an acute
angle of the one is equal to an acute angle of the other.
For they are mutually equiangular (37b, 80c, 159a).
Problem: The base of a right triangle is 12 inches long
and its altitude is 8 inches, the base of a similar rt. triangle is
15 inches long. Find the altitude.
170 GENERAL MATHEMATICS
Solution: Since the sides of similar triangles are in the
same ratio,
12 = 15
8 ~" x
Clearing of fractions
12z = 120
x = 10
Exercise
1. The sides of a triangle are 6, 7 and 8 units long. In a
similar triangle the side corresponding to 8 is 32 units long.
Find the other two sides. Hint : Form equations by aid of
Art. 158.
2. What number is in the same ratio to 4 as 10 is to 5?
160. A proportion is a statement of equality between two
equal ratios.
3 7
Thus,  = — is a proportion,
b 14
This proportion may be also written in the form 3 : 6 = 7 : 14.
This is read "3 is to 6 as 7 is to 14."
A proportion may be stated in general terms as a: b = c: d.
This is read "a is to b as c is to d."
161a. The terms of a proportion are the four quantities
compared.
Thus in the above proportion the terms are a, b y c, and d.
161b. The first and third terms of a proportion are called
the antecedents.
Thus in the above proportion a and c are antecedents.
This may also be stated as "The first term of each ratio in a
proportion is called its antecedent."
161c. The second and fourth terms of a proportion are
called the consequents.
GENERAL MATHEMATICS 171
Thus, in the above proportion, 6 and d are consequents.
This may also be stated as "The last term of each ratio of a
proportion is called its consequent."
161d. The first and fourth terms of a proportion are called
the extremes.
161e. The second and third terms of a proportion are
called the means.
These last two definitions might be combined in this state
ment: "The middle terms of a proportion are called the '
means and the terms at the two extremities are called the
extremes."
162. Theorem: In every proportion the product of the means
is equal to the product of the extremes.
Given: a: b = c: d
To Prove : ad = be
Proof: The given proportion may be written in this form:
fs < le °>
Multiplying both sides of this equation by bd,
abd cbd
"T" = T
Simplifying, ad = be. Q. E. D.
This theorem enables us to solve readily various problems
in proportion : thus,
Problem: Find the number which is in the same ratio
to 60 as 7 is to 12.
Let the number be represented by £.
Solution :
Then, 7:12 = s:60
(or a;: 60 = 7:12)
Equating the product of the means and the extremes (162),
12x = 420
Whence, x = 35.
172 GENERAL MATHEMATICS
163. The Fourth Proportional to three given quantities is
the fourth term of the proportion which has for its first three
terms the three given quantities taken in order.
Thus 8 is the fourth proportional to 3, 6, and 4 in the
proportion, 3:6 = 4:8.
By aid of (162) find the fourth proportional to:
3. 2, 3 and 4
4. 14, 15 and 28
5. 7, 9 and 21
6. 6, 35 and 42
7. 14, 15 and 16
8. 15, 16 and 14
9. 16, 15 and 14
164a. Theorem : // four quantities are in proportion, they
are in proportion by alternation ; that is, the first term is to the
third as the second is to the fourth.
Given: a:b = c:d
To Prove: a:c = b:d
Proof: From the given proportion ad = be (162).
Dividing each member of this equation by cd,
ad __ be
cd cd
Simplifying,
a _ 6
c d
Which may be written as
a:c = bid Q. E. D.
For example, this theorem might be applied to the follow
ing proportion 3 : 6 = 4:8.
That this is a true proportion may first be tested by writing
both ratios as fractions and then reducing these fractions to
lowest terms, thus,
 § thatis = 
GENERAL MATHEMATICS 173
This proportion may also be tested by equating the prod
ucts of the means and the extremes. Thus from 3:6 = 4:8
it follows that 3x8 = 4x6. That is 24 = 24.
This being an identity the proportion is correct.
By alternation 3:6 = 4:8 may be written as
3:4=6:8
164b. If four quantities are in proportion, they are in pro
portion by inversion ; that is, the second term is to the first as
the fourth is to the third.
Given: a:b = c:d
To Prove: b:a = d:c
Proof: From the given proportion be = ad (162).
Dividing each member of the equation by ac
' be __ ad
ac ac
Simplifying,
 =  or b : a = d : c Q. E. D.
a c
165. The level of still water, as the surface of a pond on a
calm day, is called the horizontal plane, or simply the
horizontal.
A line parallel to this line is called a horizontal line or
simply a horizontal.
166. A cord which suspends a weight at one end is called
a plumb line. The line determined by this cord is called a
vertical line.
167. A vertical line is perpendicular to the horizontal.
A vertical object (as for instance a flagpole) with its
shadow, forms two sides of a right triangle, the hypotenuse
of which is an imaginary line from the top of the object to
the furthest extremity of the shadow. It may be considered
as the path of a ray of sunlight bounding the shadow.
The distance of the sun from the earth is so great in com
parison with any measurement made on the earth's surface,
that we may consider the rays of sunlight as parallel lines,
174
GENERAL MATHEMATICS
without perceptible error. Thus in the following figure,
which illustrates a method of finding the height of objects
which cannot readily be measured directly, we may consider
that the hypotenuses of the two right triangles are parallel
lines.
^ *
y
T?
In the above figure, FL represents a flagpole standing on
a level plain, SL represents the shadow of the pole; PT rep
resents an upright stake short enough to be measured easily,
BT its shadow. As has just been explained, SF may be
considered parallel to BP. Then, since B T and SL are od
the same horizontal plane, Z PBT = Z FSL
since PT and FL are vertical lines,
PTB and FLS are rt. A (167)
/. PBT and FSL are similar (159b).
.. PT : FL = BT : SL
PT, BT and SL can be measured; and so the proportion
can be solved for FL (162).
10. When a fence post 5 feet long casts a shadow 6 feet
long, a flagpole casts a shadow 48 feet long. How high is
the flagpole?
11. In order to find out if the trunk of a tree below the
lowest branches is long enough to make a beam 32 feet long,
a man measures the shadow of the trunk and finds that it is
42 feet long at the same time that a fence post 4 feet high
casts a shadow 5? feet long. Is the trunk of the tree long
enough for the purpose?
GENERAL MATHEMATICS 175
XIII. Field Exercise
FINDING WIDTH OP A RIVEK
Given: In A TAB, BA is ± to TA, CE is J_ to TA and
EF is ± to AB: T, an inaccessible point.
To Derive: The value of TC.
CE is // to AB (60a)
Z T^C = Z 5PJ5A (67a)
A TEC is similar to A EBF (159b)
.•. TC : EF = CE : FB
.'. TC X FB = EF X C# (162)
Dividing both sides of the equation by FB,
EF X CE
TC =
FB
The construction of EF may be avoided by means of the
relation, „„ „ A
EF = CA (60a, 122c)
Substituting in the above,
CAX CE
iC ~~~FB
By measuring CA, CE, and FB and substituting in the
above formula, the value of TC may be computed.
Equipment: Ranging pole, half dozen stakes, tape.
Procedure: Select a location opposite a tree, or other
prominent object, on the farther bank of the river. If
possible, choose a location where the ground is smooth and
176 GENERAL MATHEMATICS
level. At some distance back from the river and directly
opposite the tree selected, drive a stake A. Walk directly
toward the tree (T) and at C drive in a stake. At C and A
lay off the J_s CD and AB. While one student stands at
B, another walks from C toward D, carrying the ranging
pole upright before him, and by direction of the observer who
is sighting f rom B to T, he locates the pole in that point on
the line CD which is in the straight line TB, marking the
position with the stake E.
Measure CE } CA, AB and the distance from C to the edge
of the river.
BA CE = BF (why?)
From these data compute the width of the river.
(The distance from C to the edge of the river is to be
subtracted from TC.)
CHAPTER XIV
FUNCTIONS OF ANGLES
168. A very important part of mathematics, called Trig
onometry, is based upon the fact that the ratios between the
sides of a right triangle depend entirely upon the size of the
acute angle, and not upon the length of the sides.
Thus, in the above figure, given that DG and BC are each
Perpendicular to AC, ADG and ABC are similar right tri
angles (159b).
/. ratio j^ = ratio rp (158).
In like manner it can be proved that the ratio of any other
perpendicular (let fall from the hypotenuse to the base as
6 F) to the base, included between the foot of the _L and the
Vertex, as FA, is also equal to the ratio of DG to AG,
I'hat is,
EF = DG
AF ~~ AG
We might proceed in this manner indefinitely, drawing
perpendiculars from the vertex and proving that the ratio
177
178 GENERAL MATHEMATICS
of the altitude to the base in ail of the right triangles thus
formed are all equal to 77; In like manner it could be
AG
shown that the ratio between the altitude and the hypote
nuse, or the base and the hypotenuse, are the same in all
these right A as they are in the right triangle ADG. Of
course, this is merely a restatement of the properties of
similar triangles, as learned in Chapter XIII, but now ap
plied to the angle.
Since A represents any given acute angle, it is evident
that for any given acute angle there is only one set of values
for the ratios between the sides of its right triangle. These
sets of values may be computed and arranged into a table
to be used in the solution of triangles.
169a. The ratios between the sides of the right triangle
of which a given angle is the acute angle, are called the
functions of the angle.
169b. A quantity is said to be a function of a second
quantity if its value depends upon the second quantity.
Thus, the total distance traveled by a train in a given
number of hours is a function of the speed of the train
that is, it depends upon the number of miles traveled by the
train in one hour. The total distance traveled is also a
function of the number of hours during which the train is
in motion.
The following ratios are called functions of an angle
because they depend entirely upon the angle.
169c. The sine is the ratio of the opposite leg to the hy
potenuse.
The cosine is the ratio of the adjacent leg to the hypote
nuse.
The tangent is the ratio of the opposite leg to the adjacent
leg.
The cotangent is the ratio of the adjacent leg to the op
posite leg.
GENERAL MATHEMATICS
Thus, in the right triangle ACB:
179
fl . A a opposite side
Snie A =  = r 7
c hypotenuse
. . A b adjacent side
Cosine A =  = r 1
c hypotenuse
A A « opposite side
Tangent A = t = ~in , .,
. 6 adjacent side
^ ± , A b adjacent side
Cotangent A =  = r: — rj
a opposite side
There are other functions of an angle, but much of the
work in the solution of triangles may be done with these four.
An important relation exists between the functions of an
angle and its complement. Thus, in the above figure the
functions of B are as follows:
Sine B = 
c
Cosine B = 
c
Tangent B = 
a
a
Cotangent B = t
o
180
GENERAL MATHEMATICS
Comparing these with the functions of A, we see that
the following relations exist between the functions of any
angle and its complement (since A and B represent comple
mentary angles in general).
170. Rule : The sine of an angle is the cosine of its com
plement, the cosine of an angle is the sine of its complement, tk
tangent of an angle is the cotangent of its complement, tk
cotangent of an angle is the tangent of its complement.
Notice that in the above pairs, one and only one function
has the prefix co (meaning "with") which indicates the
complementary relation. The relations between the func
tions of an angle and its complement may be made use of in
abbreviating a table of the numerical values of the functions
of angles, since it is only necessary to tabulate the values
from 0° to 45°, and leave the values from 45° to 90° to be
derived by aid of the above law. This has been done in
the following brief table, in which the number of angular
degrees for values less than 45° should be read at the left of
the table, and. the name of the function from the top, while
if the angle is greater than 45°, the angle should be read from
the right, and the name of the function from the bottom.
Angle
Sine
Cosine
Tang.
Cotang.
Angle
5°
10°
15°
20°
25°
30°
35°
40°
45°
.087
.174
.259
.342
.423
.500
.574
.643
.707
.996
.985
.966
.940
.906
.866
.819
.766
.707
.087
.176
.268
.364
.466
.577
.700
.839
1.000
11.430
5.671
3.732
2.748
2.145
1.732
1.428
1.192
1.000
85°
80°
75°
70°
65°
60°
55°
50°
45°
Angle
Cosine
Sine
Cotang.
Tang.
Angle
GENERAL MATHEMATICS 181
Thus, to find the cosine of 35°, we read down the column
at the left until we reach 35° — then look along this line to
the column which is marked "cosine" at the top, where we
find . 819, the function desired. Again, to find the cotangent
of 55°, we read the column of degrees at the right, beginning
at the bottom and reading up until we come to 55°, then
looking along this line until we come to the column which is
marked cotangent at the bottom, where we find .700, the
value of the function desired. Notice that we arrive at the
same number whether we look up the function of an angle,
or the complementary function of the complement of the
angle. Thus, the cosine of 35° is the sine of 55°, and the
cotangent of 55° is the tangent of 35°.
Exercise
By aid of the above table solve the following:
171. Problem: In the right A ABC given Z A = 15°, side
b = 324, find a and c.
3
Solution: First drawing the right triangle and marking
the known parts, we proceed to form equations which shall
each involve only one unknown quantity. (The functions
of known angles are considered as known quantities, since
they may be obtained by referring to the table.)
Tan A = 
b
Substituting the value of & and also substituting the value
of Tan 15° from the table,
■"*■&
182 GENERAL MATHEMATICS
Clearing of fractions
a = (.268) (324)
a = 86.832
Since the table goes only to the third decimal place
last value may be written 86.8.
In finding c we must not use an equation involving the
side just found, as to do so would be to increase the error
of our answers — since this table is correct only to three
places of decimals, that is, the value of a found is not exact,
and to use it for further computation with the tables would
increase the error: it is better to solve for the two sides
independently. Thus,
Cos A = 
c
Substituting the value of &, and also substituting the value
of cos 15° from the table,
.966 = 5*
c
Clearing of fractions, . 966c = 324
Dividing by the coefficient of c,
c = — = 335.4
.966
Since the table goes only to the third decimal place this
last value may be written 335.
These values may be checked by the formula,
c* = a 2 + b 2 (148a)
In squaring these quantities the results can be trusted to
the first three figures only: for example, 112,225 may be
written 112,000 without detracting from the accuracy of
the result.
Let the student determine what percentage the 225 so
discarded is of 112,000. Notice that the error of a square
of a quantity is greater than the error in the quantity itself,
since the error has been multiplied by the quantity.
GENERAL MATHEMATICS 183
The remaining acute angle of this triangle is found by sub
tracting A from 90°.
1. Solve the right A ABC in which A = 20°, b = 28.
(Draw a diagram as in the previous example.)
2. Solve the rt. A ABC in which A = 35°, a = 325.
3. Solve the rt. A ABC in which A = 5° and c = 284.
4. Find the area of the right triangle in which A = 25°, c = 87.
(Hint: Solve for the legs of the rt. Z and find half their
product.)
Note : Drawings of right triangles may be made to aid in
the solution of the following examples. Thus, in Ex. 5, the
horizontal distance from the observer to the building is the
base of the triangle, and the height of the building above
the observer is the altitude of the triangle.
6. Find the height of a building, if the angle of elevation
of the top is 25°, and the distance of the observer is 62 feet,
the height of the instrument being 5 feet.
(Hint : Solve the right triangle for the height of the build
ing above the observer; then add 5 feet.)
6. How long a cable is necessary to extend from a point in
an iron smokestack 48 feet above the ground and to make
an angle of 40° with the horizontal? Add 2 feet for fasten^
ing and slack of cable.
7. How high is a tree that casts a horizontal shadow 83
feet in length when the angle of elevation of the sun is 50°?
8. A ship is sailing due northeast at a rate of ten miles an
hour. Find the rate at which she is sailing due north and
also due east.
(Hint: A northeast course (line) makes an angle of 45°
with a due east course.)
9. From the top of a lighthouse known to be 141 feet above
the sea, the angle of depression of a boat is observed to be
15°. What is the distance of the boat from the lighthouse?
(Hint : The angle of depression means the angle between
the observer's line of sight and the horizontal. Hence, it is
184 GENERAL MATHEMATICS
the complement of this angle that is included in the right
triangle formed by the lighthouse, the distance along the
surface of the water from the lighthouse to the boat, and the
line of sight from the top of the lighthouse to the boat.)
In the solution of right triangles, the following rule maybe
applied to save the labor of clearing the equation of fractions:
172. Rule: With regard to either acute angle of any rigU
triangle:
Sine X hypotenuse = opposite side;
Cosine X hypotenuse = adjacent side;
Tangent X adjacent side = opposite side.
From the solution of the previous examples, the derivation
of these rules is evident. To secure facility in computation,
they should be committed to memory.
Solve for the remaining parts of the rt. A ABC, given
that :
10. A = 15°, c = 81
11. B = 20°, a = 46
12. B = 65°, c = 105
13. A = 40°, c = 91
14. When the angle of elevation of the sun is 40°, the
shadow of a tower is 170 feet long. How high is the tower?
16. In order to locate a reef the angle of elevation of the
top of a lighthouse is measured from the reef and found to
be 35°. The lighthouse is known to be 142 feet high. What
is the distance of the reef from the lighthouse?
(Hint: What is the complement of 35°?)
16. Two villages, A and B, are 7\ miles apart. A man in
a balloon directly above A observes the angle of depression
of B to be 10°. Find the height of the balloon.
17. The boom (swinging arm) of a derrick is 26 feet long,
and its lower end is fastened to the mast at a height of 3 feet
from the ground. When the angle between the derrick and
the mast is 50°, how far is the end of the boom from the
mast and how far is the end of the boom from the ground?
GENERAL MATHEMATICS 185
18. When a lighthouse is directly east of a steamer, the
captain notes the reading of the patent log (which records
the distance traveled) ; after the steamer has gone directly
north for a distance of two nautical miles, he observes the
lighthouse to lie 55° East of South. How far was the
steamer from the lighthouse at the time of each observation?
(Use 6,080 feet as the length of the nautical mile.)
XTVa. Shop Exercise
DESIGN OF RAFTER
A common instrument for making use of the properties of
imilar triangles is the Carpenter's Square — often called the
Steel Square. This consists of a flat piece of steel shaped
ike a letter L, one arm of it being usually 2 feet long and the
>ther 16 in. long, both arms being graduated in inches com
aencing at the vertex.
For example, suppose that it is desired to lay out a rafter
or a roof in which the top of the ridgepole is to be 8 feet
,bove the floor of the attic, and half the width of the attic
loor is 12 feet. Evidently we have here a right triangle,
a which the base is 12 feet, the altitude 8 feet and the
Lypotenuse is the upper edge of the rafter. If now the
Steel Square is laid on the timber to be marked in such a
ray that the 12 in. mark on the long arm coincides with the
dge of the timber, and the 8 in. mark on the short arm also
:oincides with the edge of the timber we have.a small right
riangle which is similar to the large one. (For the tangent
8 in 2
>f the acute angle at the base of the small triangle, * = ,
12 in. 3
ind the tangent of the acute angle at the base of the large
8 ft. 2
riangle equals ' = , so the acute angle at the base of
\.Zt It). o
he small triangle equals the acute angle at the base of the
arge one; therefore the triangles are similar) (159b).
186 GENERAL MATHEMATICS
The "bevel" (that is, the angle at which the mark for the I F
sawcut is to be made with the edge of the timber) both for kc
hi
m
EC
I)
k
fir
i
the top and the bottom of the rafter can be determined from
the small triangle, since the corresponding angles of similar
triangles are equal. The "foot" of the rafter may be
marked at once by drawing the pencil along the outer edge
of the long arm of the square. The problem remains to
determine the length of the rafter. (Since the right triangles
are similar, the corresponding sides are in the same ratio
that is, the hypotenuse of the small triangle will go into the
hypotenuse of the large triangle as many times as the base
of the small triangle will go into the base of the large triangle
— or as the altitude of the small triangle goes into the altitude
of the large triangle.) Accordingly the length of the hypot
enuse of the small right triangle is to be laid off 12 times
along the edge of the rafter.
Having marked the bevel along the long arm of the square,
put a pencil mark on the edge of the timber where the short I j
arm touches it (at the 8 in. mark). Move up the square
along the timber so that the 12 in. mark coincides with this
pencil mark and put a mark on the edge of the timber by
the 8 in. graduation. Repeat this process until the square
has been laid on the timber 12 times. When it is placed for
the twelfth time draw a line along the short arm of the
square for the top bevel.
It is evident that when the timber has been cut through
with a saw along these lines, and placed in position so that
the upper end is 8 feet above the horizontal, the top cut will
be vertical and the lower cut will be level.
A rafter for the roof described above would be said to have
a "run" of 12 feet and a "rise" of 8 feet. Lay out the pattern
for cutting such rafters on a smooth board 1 in. thick, 7 in.
wide and 13 feet long (1* X 7" X 13') using the method
above described.
GENERAL MATHEMATICS 187
From the outer point of the foot of the rafter measure
back 4 in. for the width of the "plate" (the timber nailed
along the edge of the attic floor to keep the rafters from slip
ping). At this 4 in. mark erect a perpendicular, lay off 2
inches along it and from this 2 in. mark draw a line (parallel
to the bottom cut) to the outer edge of the rafter. Cut out
along this line so the rafter will fit over the plate.
In a similar manner, draw a line parallel to the top bevel
to allow for the ridgepole, remembering that since rafters are
set up in pairs, it is necessary to allow only half the width
of the ridgepole on each one.
In like manner rafters could be laid out for a roof having
any given rise and run.
XlVb. Shop Exercise
DESIGN OF STAIRS
The following is the method of designing a flight of
stairs for any given position — that is, to ascend a given
height and extend a given amount from the wall, in other
words, to have a given "rise" and "run." Suppose, for
example, it was desired to lay out an extra flight of stairs to
fit in the school building.
(The application of similar right triangles to this problem
may be understood by considering a fine drawn through the
inner corners of the steps (along the stringers) erected in
Xlla. This is the hypotenuse of a right triangle in which
the line measured on the floor is the base and the fine meas
ured on the wall is the altitude. The small right triangles to
be cut from the side pieces are similar to this.)
To solve this problem, then, we state the proportion:
Rise : Run = height of riser : width of tread.
It is desirable to have the height of the riser between 8 and
9 in. It is best, therefore, to reduce all the lengths measured
to inches. The proportion then appears as follows :
Rise (in inches) : Run (in inches) = 8 \ : x.
188 ' GENERAL MATHEMATICS
If the value of x given by this proportion does not give a
satisfactory height for the riser, try some other number
instead of 8J.
To avoid such absurdities as a fraction of a step, the width
of tread must go into the run a whole number of times (and
the height of a riser the same number of times into the rise).
If possible to avoid it, the run of a flight of stairs should
not be less than the rise, or the steps will be steep.
I
CHAPTER XV
COMPUTATION OF FUNCTIONS : FOUR PLACE TABLE
OF SINES AND COSINES AND ITS USE
The values of the Functions of certain angles may be
very readily computed.
173. The functions of 60° and of 30° may be derived from
the equilateral triangle (77b, 80a), Thus:
ABC is an equilateral triangle having sides 2 units in
length. From B a JL, BD, is let fall to the base AC.
This bisects AC and the Z ABC. (Why?)
That is AD = 1 unit in length, and DC = 1 unit in length.
The Z ABD = Z DBC = 30° (since Z ABC is bisected).
Since A ABD and BDC are rt. A, the length of BD may
be computed from either of them.
BD = V AB 2  AD*
= V4  1
= V^
189
190
GENERAL MATHEMATICS
The functions of 60° may now be stated, and at the same
time those of 30° may be 'expressed by aid of art. 170.
Sin 60° =
V3
= Cos 30 c
Cos 60° = % = Sin 30°
Tan 60° =
Cot 60° =
i
i
V~3
= V~3 = Cot 30°
= Tan 30 (
The value of the sine of 60° may be found correct to four
places of decimals by extracting the square root of 3 to five
figures and then dividing this result by 2. In like manner
the other functions of 60° and the functions of 30° may
be worked out.
174. The values of the functions of 45° may be computed
from the isosceles right triangle. Thus:
In the rt. A ABC, leg AC = leg BC.
:. Z BAC = Z ABC (77b).
Then, since Z C is a rt. Z , Z BAC = 45°, and Z ABC
45° (80a).
Let AC be 1 unit in length, and BC be 1 unit in length.
GENERAL MATHEMATICS 191
The length of AB may now be obtained as follows:
AB* = AC* + BC*
AB = j/ AC* + W*
= {/ 1 +1 = V2
The functions of 45° may now be expressed as follows:
Sin 45° = I
V2
Cos 45° = ^
V2
Tan 45° =  = 1
Cot 45° = j = 1
Notice that sin 45° = cos 45° and tan 45° = cot 45°.
Since 45° is its own complement, this agrees with art. 170.
The functions of different angles may be obtained by the
aid of the protractor — laying off the angle on the paper
carefully with a ruler and hard pencil — then drawing a
perpendicular to one side of the angle from some point on
the other side. The sides of the right triangle thus formed
may now be measured with the ruler.
It is usually better not to take the inch as the unit of
measure, but some smaller division, such as the eighth of an
inch or the sixteenth. If the same unit is taken for all the
sides, it is not necessary to write the lengths of the sides in
the fractional form, but simply to write the number of units
in each side. Thus, suppose the sides of a triangle are
21 28 35
— > —f and — long — the ratios will be the same if the sides
16 16 16
are written as 21, 28, and 35.
192 GENERAL MATHEMATICS
Problem :
Find to four places of decimals the value of the functions
of the Z A, given that a = 17, and b = 16, in the right A ABC.
Solution: c = Va 2 + 6 2
= V(17) 2 + (1 6)*
= V289 + 256 = ^545 = 23.345
••• Sin A  2^45  7282
°" A s!iffi •«*
Tan = ^ =1.0625
16
Cot A = j = .9412
Exercise
In the right triangle ABC find the functions of the Z i,
given that,
1. a = 7, 6 = 18
2. a = 34, c = 41
3. 6 = 28, c = 39
4. a = 17, c = 37
Draw the above right triangles carefully to scale and
measure the angle with a protractor.
Lay off the following angles with the protractor, using
1/16 of an inch as a unit of measure; measure off any con
venient length and complete the right triangle ; measure the
remaining sides and compute the values of the functions:
5. 6. 7. 8.
32° 41° 76° 54°
By making the right triangles sufficiently large, the func
tions of any angle may be computed to the degree of accuracy
GENERAL MATHEMATICS
193
desired. In making a table, however, it is best to compute
the functions algebraically (as was done in obtaining the
functions of 30°, 45° and 60°) from a formula, of which there
are a great number.
The following table of aines and cosines has been computed
from formulas and is accurate to the fourth decimal place.
175a. Natural Sines and Cosines.
z°
Sine
Cosine
Z°
Sine
Cosine
1°
.0175
.9998
89°
24°
.4067
.9135
66°
2°
.0349
.9994
88°
25°
.4226
.9063
65°
3°
.0523
.9986
87°
26°
.4384
.8988
64°
4°
.0698
.9976
86°
27°
.4540
.8910
63°
5°
.0872
.9962
85°
28°
.4695
.8829
62°
6°
.1045
.9945
84°
29°
.4848
.8746
61°
7°
.1219
.9925
83°
30°
.5000
.8660
60°
8°
.1392
.9903
82°
31°
.5150
.8572
59°
9°
.1564
.9877
81°
32°
.5299
.8480
58°
10°
.1736
.9848
80°
33°
.5446
.8387
57°
11°
.1908
.9816
79°
34°
.5592
.8290
56°
12°
.2079
.9781
78°
35°
.5736
.8192
55°
13°
.2250
.9744
77°
36°
.5878
.8090
54°
14°
.2419
.9703
76°
37°
.6018
.7986
53°
15°
.2588
.9659
75°
38°
.6157
.7880
52°
16°
.2756
.9613
74°
39°
.6293
.7771
51°
17°
.2924
.9563
73°
40°
.6428
.7660
50°
18°
.3090
.9511
72°
41°
.6561
.7547
49°
19°
.3256
.9455
71°
42°
.6691
.7431
48°
20°
.3420
.9397
70°
43°
.6820
.7314
47°
21°
.3584
.9336
69°
44°
.6947
.7193
46°
22°
.3746
.9272
68°
45°
.7071
.7071
45°
23°
.3907
.9205
67°
Cosine
Sine
Z°
Cosine
Sine
Z°
In this table, the number of degrees in an angle less than ,
45° is read from the left hand column, and the name of the
194 GENERAL MATHEMATICS
function read at the top; the number of degrees in an angle
greater than 45° is read from the right hand column, and the
name of the function read at the bottom.
Problem: Find the sine of 37°.
Solution: Since this angle is less than 45°, we read down
the column of figures at the left until we find 37° ; the func
tion opposite it ( . 6018) is the function desired, that is, we
read the function from the column adjoining the column of
degrees just read, as this column of functions is labelled
"Sine" at the top.
Problem: Find the sine of 78°.
Solution : Since this angle is greater than 45° we read the
angle from the column at the right, reading upward until we
reach 78°. The function next it, on the same line (.9781),
is the function desired — since this function is labelled "Sine"
at the bottom.
Problem: Find the cosine of 27°.
Solution: Since the angle is less than 45° we read down
the column of degrees at the left until we encounter 27°;
the function on the same horizontal line with 27°, but in the
second column from the number of degrees just read, (.8910)
is the function desired, since this column is labelled "Cosine"
at the top.
Problem : Find the cosine of 64°.
Solution : Since this angle is greater than 45° we read the
angle from the column at the right, reading upward until we
reach 64°. The function on the same horizontal line with
64°, but in the second column from the number of degrees
just read, (.4384) is the function desired, since this column
is labelled " Cosine' ' at the bottom.
From the above examples the following rule appears with
reference to the preceding table:
GENERAL MATHEMATICS 195
175b. The sine of a given number of degrees is thefunctiort
on the same horizontal line, next adjoining the number of de~
grees : the cosine of a given number of degrees is the next adjoin
ing function but one, on the same horizontal line.
Problem: Find the sine of 58° 25'.
Solution: Sin 59° = .8572
Sin 58° = .8480
Tabular Difference .0092
Since . 0092 corresponds to an increase of 1° in the angle,
25' will correspond to the same fraction of .0092 as 25' is
of 1°. Ttas, 25'  (I)' = ( I)'
^ X .0092 =  X .0023 = ^~ = .0038
Sin 58° = .8480
Correction = .0038
Sin 58° 25° = .8518
Since the sine is an increasing function (that is, the sine
increases as the angle increases), this correction is added.
After finding a function corresponding to a given number
of degrees and minutes, the student should check his work
by noticing that it falls between the proper two functions
in the table. Thus,
Check : This value lies between the sine of 58° and of 59°
as given by the table.
175c. The change in a function corresponding to 1° is
called the Tabular Difference. (See above problem.)
176a. To find the sine corresponding to a given number of
degrees and minutes: drop the minutes and find from the table
the sine corresponding to the given number of degrees ; subtract
this from the next greater sine in the table to obtain the tabular
difference; express the given number of minutes as a fraction
of a degree, multiply the tabular difference by this fraction, and
add this product to the sine first obtained from the table.
196 GENERAL MATHEMATICS
Exercise
Find the sines corresponding to the following angles:
1. 2. 3. 4. 6.
13° 27' 49° 48' 21° 35' 56° 45' 37° 40'
6. 7. 8. 9. 10.
38° 25' 72° 42' 78° 15' 81° 24' 82° 18'
The work of finding the cosine corresponding to a given
number of degrees and minutes is like that of finding the
sine except that the correction is subtracted instead of added,
since the cosine is a decreasing function — that is, it decreases
as the angle increases.
Problem: Find the cosine of 74° 54'.
Solution: Cos 74° = .2756
. Cos 75° = .2588
Tabular Difference .0168
M# (S)° (ro)°=^ x  0168 =  01512
Cos 74° = .2756
Corr. = .0151
Cos 74° 54' = .2605
Check : This value lies between the cosine of 74° and the
cosine of 75° as given by the table.
176b. Rule: To find the cosine corresponding to a given
number of degrees and minutes: drop the minutes and find
from the table the cosine corresponding to the given number of
degrees; from this subtract the next less cosine in the table to
obtain the tabular difference; express the given number of
minutes as a fraction of a degree, multiply the tabular difference
by this fraction, and subtract this product from the cosine first
obtained from the table.
GENERAL MATHEMATICS 197
Find the cosines corresponding to the following angles:
' 11. 12. 13. 14.
28° 35' 37° 42' 49° 12' 51° 20'
15. 16. 17. 18.
63° 36' 10° 45' 41° 10' 44° 24'
Problem : Find the angle corresponding to the sine . 6738.
Solution : We first locate the given sine between the next
less and the next greater sine in the table; thus,
Sin 43° =.6820 Given Sin =. 6738
Sin 42° =.6691 Sin 42° = .6691
Tab. Dif . = .0129 Given Dif . = . 0047
Since .0129 corresponds to a difference of 1°, the Given
Difference will correspond to that fraction of 1° which .0047
is of .0129. It is best to change this ratio '  to a deci
mal; thus,
.0129 1 .0047 1 .36
387
830
The correction to be added to 42° is therefore .36 of a
degree. Changing this to minutes
.36
60
21.60'
or to the nearest whole minute, 22'. The angle is therefore
42° 22'.
177a. Rule : To find the angle corresponding to a given sine:
locate the given sine between the next greater and the next
less sine in the table; find the decimal part which the given
difference is of the tabular difference, multiply this by 60 and
add this product as minutes to the angle corresponding to the
smaller sine.
198
GENERAL MATHEMATICS
Exercise
Find the angle whose sine is:
1. 2. 3.
.4291 .6843 .3471
6.
.8260
7.
.7452
8.
.2100
4.
.5893
9.
.1564
6.
.9431
Since the cosine decreases as the angle increases, in
finding the angle corresponding to a given cosine, the correc
tion must be subtracted. Thus,
Problem: Find the angle whose cosine is .5906.
Solution : We first locate the given cosine between the next
less and the next greater cosine in the table:
Cos 53° = .6018 Given Cos = .5906
Cos 54° = . 5878 Cos 54° = .5878
Tab. Dif. = .0140 Given Dif. = .0028
Since .0140 corresponds to 1°, .0028 will correspond to
that fraction of a degree which .0028 is of .0140. It is best
to change this ratio, ' > to a decimal; thus,
.0140
.0140  .0028
280
L?
Changing this to minutes,
60
.2
12.0'
54°  12' = 53° 48' Ans.
177b. Rule: To find the angle corresponding to a given
cosine: locate the given cosine between the next greater and the
next less cosine: find the decimal part which the given differ
ence is of the tabular difference, multiply this by 60 and sub
tract this product as minutes from the angle corresponding to
the smaller codne.
GENERAL MATHEMATICS 199
ind the angle whose cosine is:
10.
n.
12.
18.
14.
.6942
.3681
.5724
.8702
.2073
16.
16.
17.
18.
19.
.7005
.4506
.3972
.8291
.1246
l becoming accustomed to the use of these rules it is
ssistance to reflect: "How shall the correction be dealt
l so that the angle shall fall between the two taken from
table?" Since in the case of the Sine (the increasing
tion) the correction is added, it must be added to the
Her of the angles taken from the table so that the angle
I lie between those two angles: whereas, in the case of
Cosine (the decreasing function) it must be subtracted
i the larger so that the answer may fall between those
angles — since to add to the larger or subtract from the
ller would make the answer fall outside.
Ithough the established custom of giving problems in
ing seconds gives the student the idea that a second of
ee or arc is an important unit of measure, it can often
is be neglected in practical problems without sensible
r.
aking the circumference of the whole earth as 25,000
s, let the student determine the number of miles in 1°,
number of miles in 1', and then the number of feet in 1".
e the length of arc corresponding to 1" is so small
aparatively) when the radius is 4000 miles, the student
understand why the second is neglected in these prob
»•
r 8a. North, East, South and West are called the Cardinal
Lts of the Compass.
r 8b. The direction of a line as shown by the compass is
d its bearing.
200 GENERAL MATHEMATICS
The bearing of a line which does not coincide with any
of the cardinal points, may be indicated by stating the num
ber of degrees it is East of North or West of North; also by
the number of degrees it is East of South or West of South:
thus, a line which makes an angle of 7° on the West side of the
North and South line — going in a direction approximately
South would be marked as S. 7° W. Of course, if the ob
server reversed his direction of movement and proceeded
along the same line in the opposite direction, he would be
going N. 7° E.
Gunter's Chain, used in compass surveying, is divided
into a hundred links. Hence, if a side of a field was 3 chains
and 41 links long it might be written as 3 . 41 ch. As has
previously been explained (Chap. IX), to change square
chains to acres (sq. ch. to A.) divide by 10 (see 129).
Problem : Starting from a point A a surveyor runs N. 26°
35' E. 9 . 80 chains to B, thence N. 56° 15' E. 8 . 10 chains to C,
thence S. 5° 30' W. 24. 10 chains to D, thence N. 39° 30' W.
13.90 chains to A. Calculate the area of the field A BCD.
Solution : Draw a diagram of the field to scale, using the
protractor to lay out .the angles (remember that the top of
the page is north, the right hand east, etc.) At the left
side of thedrawing draw a North and South line. On this
line locate the most westerly point, A. Lay off AB. To
assist in laying off BC, a N. and S. line BM may be drawn
through jB, and from this an angle of 56° 15' laid off to the
right. When BC and CD have been laid off, the drawing
may be checked by connecting C and D and then measuring
the length of the line and its bearing, to see if they are of the
value stated. Note that the Z EAD is the same as the
bearing of DA — that is, the same as the angle which AD
would make with an N. and S. line drawn through D (64).
From the diagram it appears that,
(I) Area ABCD = area trapezoid GCDE — (A ADE +
A ABF + trapezoid FBCG).
le following
lod of solu
. not only
3 the answer
checks the
iracy of the
ey. Certain
terms must
be defined:
Tien the sur
:>r goes from
> B, the final
tion of B is
same as if
had moved
a A to F and
1 from F to
The distance
red directly
th (as AF)
sailed the
rthing and
3 distance
mi directly
b (as FB) is
ed the East
• Likewise
T is a North
andMCisan
ting. Again
en the sur
ror moves
nCtoD,the
,1 location of
3 the same as
GENERAL MATHEMATICS
No**
201
202
GENERAL MATHEMATICS
if he had moved from C to P and thence from P to D. The
distance moved directly south (as CP) is called the South
ing, and the distance moved directly west (as PD) is called
the Westing. Note that Northing is the opposite of South
ing and Easting is the opposite of Westing.
Obviously, if the surveyor goes all around the field and
returns to the same place from which he started, for the dis
tance which he went north he must return an equal distance
toward the south, and for the distance which he went east,
he must return an equal distance west.
178c. Rule: // a survey completely around any area is
correct, the sum of the northings equals the sum of the southings,
and the sum of the eastings equals the sum of the westings.
In solving for areas of (I), the work may be arranged in
the following manner to test the accuracy of the survey.
Note: To understand why the line EA is computed in the following
solution, and the angle EAD used instead of Z QDA (the bearing of the line),
draw a N. and S. line through D, and an E. and W. line through A. Let Q
be their point of intersection. Evidently QD = AE, AQ = ED (60a, 122c),
Z QDA = Z EAD (64).
AF = AB Cos 26° 35' = ( 9.80) (.8942) =
BM = BC Cos 56° 15' = ( 8.10) (.5555) =
EA = AD Cos 39° 30' = (13.90) (.7715) =
CP = CD Cos 5° 30' = (24.10) (.9953) =
FB = AB Sin 26° 35'
MC = BC Sin 56° 15'
DP = DC Sin 5° 30'
ED = AD Sin 39° 30'
( 9.80) (.4475)
( 8.10) (.8314)
(24.10) (.0959)
(13.90) (.6361)
N.
S.
8.76
4.50
10.72
23.99
23.98
23.99
E.
W.
4.39
6.73
2.31
8.84
11.12 11.15
As will be seen from the above, the Northings do not exact
ly balance the Southings, nor the Eastings the Westings — but
GENERAL MATHEMATICS 203
the error is so small that the survey may be classed as accu
rate. It is unusual, in the survey of an area of this size, to
have any smaller error than the above. If there is a large
discrepancy, however, the survey should be repeated. The
error in the area is reduced by the above method of solution;
since one area is subtracted from another.
Let the student compute the maximum error in the above
computation, in percentage. (Hint: What per cent is 3 links
of 11.12 chains?)
178d. Northings and Southings are called Latitudes, and
Eastings and Westings are called Departures.
By aid of the latitudes and departures, the areas in equa
tion (I) may be computed.
Area A ADE = £ (EA X ED) =
J (10.72 X 8.84) = 47.38 sq. ch.
Area A ABF = £ (FA X FB) =
J (8.76 X 4.39) = 19.23 " "
Area Trapezoid FBCG = \ (GC +
FB) X BM = i (15.51 X 4.50) = 34.90 " "
101 . 51 sq. ch.
Area GCDE = % (GC + ED) X CP
= \ (19.95 X 23.99) = 239.42 sq. ch
Minus areas = 101.51 " "
Area ABCD = 137.91 sq. ch.
or 13.79 Acres.
In the above computation GC is found by adding FB and
MC. If desired, instead of dividing each different product
in the above by 2, as in the above solution, which takes
advantage of cancellation, this division may be left until the
last, as shown by the following:
 [ (GC + ED) X CP]  J [EA XED]  § [FA X FB]  *
[ (GC + FB) X BM] =
J J [ (GC + ED) X CP]  [EA X ED]  [FA X FB] 
[ (GC + FB) X BM]
204
GENERAL MATHEMATICS
Exercise
1. A deed of a woodlot gives the following boundaries
(the letters A, B, C and D are used instead of the description
of the corners as given in the deed) : "Starting from a point
A, N. 31° 30' W. 10.40 chains to B y thence N. 62° E. 9.20
chains to C, thence S. 36° E. 7. 60 chains to X>, thence S. 45°
30' W. 10 chains to A." Find the area of the woodlot.
2. A surveyor, starting at A, measures S. 50° 30' E. 6 chains
to B, thence S. 58° W. 4.20 chains to C, thence N. 28° 10'
W. 5 chains to D. What is the bearing of A from D. and
how long is AD?
(XV) Finding Area of Field by Tbiangulation:
Equipment: Transit (horizontal angle instrument), two
ranging poles, measuring tape, stakes.
..^
«•*'
E^ k
MHM
Procedure : At the edge of the field most suitable, measure
a base line AB. In case it is not possible to measure this
base along the edge of AB directly (due to a ditch, or
GENERAL MATHEMATICS 205
growth of bushes, or similar obstruction) this line may be
laid off to one side of AB and parallel to it (as the line
A'B'); and the length of AB projected upon it, as CD.
As the computation of all the triangles will depend upon
the base line AB, this line must be measured with consider
able care. The transit should be placed at one end of
the line A'B', and by aid of the ranging pole a straight line
should be laid off as in Field Exercise II. Then perpendicu
lars should be constructed from A to A'B', and from B to
A'B', and the distance CD, between the feet of these per
pendiculars, should be carefully measured.
The distance AB being known, the transit should be
set up at one end as A, and a ranging pole set up at the other
end B. The other ranging pole should be set up at some
suitable place, E, near the center of the field, which should,
if possible, be visible from all four corners of the field.
Measure Z BAE. Move the transit to B and meas
ure the Z ABE. (Since one side and two angles of the
triangle are known, the triangle can later be solved by aid
of Chapter XVII.) In like manner set up the instrument at
B and measure Z EBF. Then move the instrument to E
and measure BEF. Since the side BE will be known when
the A AEB is solved, it is possible to solve A BEF, since
a side and two adjacent angles are known.
The A EFG will be solvable if A GEF and EFG are known
since the side EF can be found by solving A FEB.
Likewise the A AEG will be solvable if A AGE and AEG
are known since the side GE can be obtained by solving
A GEF.
As a check on the accuracy of the work, all the angles in
the diagram should be measured, when any error in observa
tion may be detected by noting if the sum of the angles of
each triangle is equal to 180°, and also if the sum of the angles
about the point E is equal to 360°.
In making any correction in the observations, try to
206 GENERAL MATHEMATICS
"distribute the error" proportionally. For example suppose
the three angles measured in the triangle AEG are 34° 5',
75° 21' and 70° 36'. Since the total of these amounts to
180° 2', it is necessary to subtract 2' from this. From the
largest Z, 75° 21', subtract 1', from the second largest Z,
70° 36', subtract 1', leaving the' smallest angle uncorrected.
(A correction of 1' in an Z of 75° 21' is much less, propor
tionally, than a correction of 1 ' in an angle of 34° 5'.)
The corrected readings of the three angles, then, are
34° 5', 75° 20' and 70° 35', giving a total of 180°.
The solution of these triangles is to be left until after
studying the Sine Law (Chap. XVII).
(For a large field, in which there are many obstructions to
the view, it will be necessary to establish other centers of
angles besides E. )
CHAPTER XVI
RELATIONS BETWEEN THE FUNCTIONS: AXES OF
REFERENCE: FUNCTIONS OF OBTUSE ANGLES
179a. Theorem: The sine of any angle divided by its cosine
equals the tangent of that angle.
Given: A, any acute angle.
To Prove: S in A
= Tan A
Cos A
Proof: Complete the rt. A ABC
Tan A = ?
a
(169c)
Since t is a fraction, both of its terms may be divided by
the same quantity without changing its value. Hence,
a
a)
207
208 GENERAL MATHEMATICS
But  = Sin A and  = Cos A (lQ9c).
c c
Substituting these values in (I),
Tan A = ^A Q. E. D. •
Cos A
By clearing this equation of fractions,
179b. Tan A Cos A = Sin A
Note: The proof for the case when Z A is obtuse may be worked out
after studying the theorems on the functions of obtuse angles, given in the
latter part of this chapter.
Exercise
1. Check the tangent of the following angles, as given in
the table of Chapter XIV, by dividing the sine by the cosine:
(a) (b) (c) (d) (e)
20° 65° 50° 70° 15°
2. Check the sine of the following angles by multiplying
the tangent by the cosine:
(a) (b) (c) (d) (e)
10° 35° 55° 70° 50°
3. By aid of the table in Chapter XV, derive to four places
of decimals the tangent of the following angles:
 , (a) (b) (c) (d)
31° 47° 38° 15' 11° 24'
4. Prove that Cot A = = — T
Sin A
Hint: Proof is similar to that of 179a.
180a. Unity means the numeral one. This term is used
to avoid confusion when one is used abstractly (that is, with
out qualifying any object) as in the following theorem.
GENERAL MATHEMATICS
209
180b. Theorem : The square of the sine of any angle, plus
the square of the cosine of that angle, equals unity.
Given : L A any acute Z .
To Prove: (Sin A) 2 + (Cos A) 2 = 1
Proof: Complete the right triangle, BAC.
a 2 + b 2 = c 2 (Pythagorean Theorem)
Dividing each term of this equation by c 2 ,
n 2 7)2 c %
(I)  +  =
»2
But
c 2 \c/
6* = /&y
c 2 \c/
Substituting in I,
(ID
But  = Sin A, and  = Cos A
c c
Substituting in II,
(Sin A) 2 + (Cos A) 2 = 1.
180c. (Sin A) 2 may be written Sin 2 A:
(Cos A) 2 may be written Cos 2 A.
210 GENERAL MATHEMATICS
The above theorem may then be conveniently expressed
by this formula:
180d. Sin*A + Cos'A = 1.
The above formula may be solved for the cosine by trans
posing Sin 1 ; thus:
Writing the equation in the general form,
Sin 1 + Cos* = 1.
Transposing, Sin 1 = 1 — Cos 1 .
Extracting the square root of both sides,
180e. Sin = Vi _ Cos*
In like manner, by solving for the cosine,
180f. Cos = V l  Sin*.
180g. From the two preceding formulas it is evident that
the value of the sine is less than unity, since the sine is
equal to the square root of unity minus a quantity. For
the same reason, the value of the cosine is less than unity.
This is evident, also, from the consideration of the fact that
the sine is one leg of a right triangle divided by the hypot
enuse, and the cosine is the other leg of a right triangle
divided by the hypotenuse, for since the hypotenuse of a
right triangle is longer than either leg, the numerator of each
of these fractions is less than the denominator.
A further consideration of these formulas shows that as
the sine increases the cosine diminishes. And as the sine ap
proaches unity in value, the cosine approaches zero (nothing)
in value.
These results may be summed up in the following:
180h. Rule : The value of the sine of an angle lies between
zero and unity.
180i. Rule : The value of the cosine of an angle lies between
unity and zero.
The above formulas may be used to derive functions or to
test the accuracy of functions which have been already
derived. Thus,
GENERAL MATHEMATICS 211
ven: Sin 37° = .6018; find Cos 37°.
lution :
Cos 37° = V 1  (Sin 37° )*
= v l  (.6018) 2
= v 1  .362 16324
= v . 63783676
= .7986
Exercise
Given: Sin 48° = . 7431, find Cos 48° by above formula
compare with value given in the table.
Given: Cos 39° = . 7771, find Sin 39° by above formula
3ompare with the value given in the table.
Given that Cos 28° = .8829 and Sin 28° = .4695, test
accuracy by squaring both functions and adding,
r sum should equal 1.
If the sine of an angle, obtained as the answer to some
lem, turns out to be in the form 1 . 6789, why should the
: be repeated? (180h).
Find Sin 19° 30' and Cos 19° 30'. Check as in exam
I.
far the functions considered have been those of angles
than 90°, but it is possible to extend the ideas of func
i to include angles greater than a right angle. For
ng with such angles the following system has been
d convenient:
Let the student prove that the product of the tangent
l angle by the cotangent is unity.
What is the product of the tangent and cotangent of
is determined from the table on page 180?
From the same table determine whether or not the value
le tangent and cotangent are restricted in the same
ler as the sine and cosine (180h, 180i).
212
GENERAL MATHEMATICS
181a. A plane may be considered as being divided by two
intersecting straight lines called axes, intersecting each other
at right angles. Thus,
181b. The axis parallel to the top of the page is called the
axis of X, or Xaxis; the axis parallel to the side of the page is
called the axis of Y, or 7axis.
181c. The point of intersection of these two axes is called
the origin.
As any point in a plane may be considered as the origin,
it will be possible to derive the functions of any angle by
referring them to an origin at the vertex of the angle — that is,
through the vertex of any angle in a plane we may draw two
straight lines intersecting at right angles.
181d. These axes divide a plane into four parts called
quadrants.
XOY is called the First Quadrant, YOX' the Second
Quadrant, X'OY' the Third Quadrant, Y'OX the Fourth
Quadrant.
The Xaxis and yaxis together are called the axes of
reference.
181e. The perpendicular distance from any point in the
plane to the Xaxis is called the ordinate of the point; the
distance measured along the Xaxis from the foot of this
ordinate to the origin is called the abscissa.
GENERAL MATHEMATICS 213
Thus, in the above figure, OF is the abscissa of the point
P and PF the ordinate.
Any point in the plane may be located by stating the length
of its abscissa and ordinate.
181f . The abscissa is always written first.
Thus, the point (5, 3) means the point whose abscissa is
5 and whose ordinate is 3.
If the student attempted to lay out this point without
further instruction, he would find four points in the plane as
the point (5, 3), thus: measuring 5 units to the right of the
origin, and then 3 above the Xaxis, he would locate a point
iti the first quadrant ; producing this ordinate for a length of
3 below the Xaxis he would locate another point in the
fourth quadrant; then measuring 5 to the left of the origin
and 3 above the Xaxis he would locate another point in
the second quadrant; producing this ordinate below the line
3 units he would locate another point in the third quadrant.
This ambiguity is removed by observing the following:
18 lg. Rule: Abscissas measured to the right of the Yaxis
are positive, to the left negative; ordinates measured above the
Xaxis are positive, below negative.
Marked according to this method, the point in the first
quadrant would be the point (5, 3), that in the second, the
point (  5, 3), that in the third the point (  5,  3) and
that in the fourth the point (5,  3).
181h. The abscissa and ordinate together are called the
coordinates of a point.
Evidently a point is definitely located when its coordi
nates are stated 1 — as there is only one point in a plane which
has a given abscissa and ordinate.
Note : A similar system is used for locating places on the earth's surface,
in which the longitude gives the location of a place east or west of the meridian
of Greenwich, and the latitude gives its position north or south of the equator.
The meridian of Greenwich corresponds to the Faxis, the equator to the X
axis, and their intersection to the origin.
214
GENERAL MATHEMATICS
Exercise
Lay off an X and a Y axis (at right angles to each other")
and using a half inch as a unit of measure, locate the follow
ing points:
(a) (6,1) (b) (3,7) (c) (5, 1) (d) (5, 2)
(e) (6, 3) (f) (2,1) (g) (3,8) (h) (7, 3)
These rules may now be applied to stating the functions of
angles. To avoid ambiguity, it is customary to observe the
following:
182a. Rule : An angle begins with the OX line and moves in
the contrary direction to the hands of a watch.
182b. The OX line is caUed the initial line. The other
boundary line of the angle is called the terminal line.
182c. The distance is the hypotenuse of the right triangle
of which the abscissa is the base and the ordinate is the
altitude. The distance is always positive.
Note : The distance is always part of the terminal line.
Thus, in the accompanying figure, OX is the initial line,
OC is the abscissa, PC is the ordinate, and OP is the dis
tance. From this figure we may derive the following:
182d. The sine is the ratio of the ordinate to the distance.
The cosine is the ratio of the abscissa to the distance.
The tangent is the ratio of the ordinate to the abscissa.
The cotangent is the ratio of the abscissa to the ordinate.
GENERAL MATHEMATICS
215
Denoting the abscissa by x, the ordinate by y, and the
distance by d, we have the following:
Sin Z POX = \
a
Cos Z POX = \
a
Tan Z POX = ^
x
Cot Z POX = 
1/
From an inspection of these ratios we may derive the fol
lowing:
182e. Rule : All the functions of an angle in the first quad
r dnt are positive.
In a similar manner, we may derive the functions of an
angle in the Second Quadrant, that is, of an angle greater
than 90°, but less than 180°.
Thus, in the accompanying diagram, the angle POX has
been generated by the revolution of the terminal line, about
216
GENERAL MATHEMATICS
0, from the position OX to the position OT. The angle
XOT is evidently greater than the right angle XOY.
To compute the functions of Z XOT, from any point
on TO, as P, a perpendicular is let fall to the Xaxis, there
by forming the right triangle PCO. The functions of the
angle POC may now be read off as follows:
Sin Z POC =
if
<2
Cos / POC =
—a;
T
=
X
Tan Z POC =
—X
=
_ 2/
a;
Cot Z POC =
a;
2/
=
y
From an inspection of these ratios we may derive the fol
lowing:
182f . Rule : The sine of an angle in the second quadrant is
positive; its cosine, tangent, and cotangent are negative.
This rule enables us to derive a relation between the func
tions of an angle and those of its supplement. Thus, in the
accompanying diagram:
t; y
T"
P <
;*
r
GENERAL MATHEMATICS 217
en: Z TOX an angle in the Second Quadrant, P any
on its terminal line, and PD and OD its ordinate and
sa.
Derive: Functions of Z TOX in terms of its supple
ivation: On OX, with as a vertex, construct the
T'OX (in the First Quadrant) equal to TOX'. Lay
*' equal to OP, and from P' let fall a perpendicular
i rt. A PDO = the rt. A P'OD' (Why?)
/. P'D' = PD
0D f = OD
0P f = OP
t line OP, and its equal OP' may be represented by d>
nd its equal P'D' by j/, and OD by  a; and OD' by x.
3 evident from the above diagram that TOX' is the
jmentof TOX.
\ functions of T'OX (the equal of TOX') will there
b the functions of the supplement of TOX. Z T'OX
e referred to as "Supplement Z TOX. )y From the
diagram, we may, by inspection, obtain the following
TOX = \ Sin (Supplement Z TOX) = ^
a a
TOX =~ =  5 Cos (Supplement Z TOX) = ~
ad a
: TOX = £ =  % Tan (Supplement Z TOX) = ^
— £ £ X
TOX = — =   Cot (Supplement Z TOX) = 
y v y
218 GENERAL MATHEMATICS
By comparison of these two tables of functions we may
obtain the following:
182g. Rule : The sine of an angle is equal to the sine of its
supplement; the cosine of an angle is equal to minus the cosine
of its supplement; the tangent of an angle is equal to minus the i
tangent of its supplement; the cotangent of an angle is equal to
minus the cotangent of its supplement.
This rule enables us to find the functions of an angle greater
than 90°, although the table reads only to 90°. For example,
to find the sine of 131° 20' we look up the sine of 48° 40'.
Again, to find the cosine of 127° 15' we look up the cosine
of 52° 45' and write it with a minus sign in front of it.
To find the angle corresponding to the cosine .3781 we
look up the angle corresponding to .3781 and then find the
supplement of this angle, since the minus sign indicates that
the angle is greater than 90°.
On the other hand, when we are required to look up the
angle corresponding to a given sign, it is impossible to deter
mine from the function whether we should use the angle
which we take from the table (an acute angle) or its sup
plement, an obtuse angle.
In solving a triangle, if only one of the angles is given,
this peculiarity of the sign requires the application of special
rules. (Chap. XVIII.)
Exercise
1. Find the sine of the following angles:
(a) 108° 10' (b) 124° 25' (c) 162° 50'
(d) 118° 45' (e) 121° 35' (f) 134° 40'
2. Find the cosine of the following angles:
(a) 126° 15' (b) 149° 40' (c) 152° 30'
(d) 109° 24' (e) 115° 48' (f) 161° 35'
3. Let the student prove that the theorems of 179a and
179b hold true for obtuse angles.
4. Let the student prove that the formulas of 180d, 180e
and 180f hold true for obtuse angles.
GENERAL MATHEMATICS
219
For comparing the rate at which the functions change as
the angle changes, it is convenient to take the terminal line
as of fixed length — that is, as it revolves about the origin
to consider its extremity as describing a circle, as in the
accompanying diagram :
Let the terminal line, as it revolves counterclockwise
about the origin, from OX take the successive positions,
PO\ P'O, P"0, and let PD, P'D', and P"D" be the perpen
diculars let fall from the points P, P', and P", which are
points on the circumference of a circle.
Therefore :
Sin Z POX =
PD
OP
Sin Z P'OX =
Sin Z P'OX =
P'D'
OP'
P f 'D"
OP"
220 GENERAL MATHEMATICS
Since OP = OP' = OP\ they may be replaced by r
(radius); these ratios may then be. written as follows:
PD
Sin Z POX =
r
Sin L P'OX =
P'D'
r
Sin L P'OX =
P'D'
By inspection it can be seen that as the angle increases the
value of the numerators of these fractions increases, while
the denominators remain constant in value — that is, the
value of the fraction increases. Likewise:
Cos Z POX = —
Cos Z P'OX =
Cos £ P'OX =
r
OP'
r
OD"
By inspection it can be seen that as the angle increases the
value of the numerator of these fractions decreases, while
the denominator remains constant — that is, the value of the
fraction decreases.
Note: If desired, this might be rendered more obvious
by drawing the angles so that the lines could be exactly
scaled. Thus, the successive values might be* written:
Sin Z POX = jjj Sin Z P'OX = ^
Sin Z P'VX = ^
GENERAL MATHEMATICS
221
This makes it evident that as the angle increases the sine
increases.
Likewise (by drawing another figure),
Cos Z POX = j
Cos Z P'OX =
10
Cos Z P'OX = ^
These results may be summed up in the following:
183a. Rule : As the angle increases from 0° to 90°, the sine
increases while the cosine decreases.
Let the student compare this rule with arts. 177a and
177b.
Exercise
1. Check the accuracy of the above rule by tracing the
successive values of the sine and cosine from 1° to 89° in
the table.
2. Construct a diagram and prove that this rule is re
versed for angles in the second quadrant.
r*
222 GENERAL MATHEMATICS
This method of comparing the values of the functions of
angles by considering the terminal line as of definite length,
brings out another important fact.
When the terminal line swings from OX to the position
OP, describing a small angle, the movement of its outer
extremity approximates a perpendicular to the Xaxis— that
is, its ordinate increases very rapidly, while its abscissa de
creases very slightly. In other words, for very small angles,
the sine changes in value very rapidly, while the cosine
changes very slowly. For angles near 90° this is reversed.
Let the student verify this statement by comparing the
change for 1° in these functions for angles less than 5° with
the changes for 1° with some angles near 45°.
Evidently, if any error has been made in observing the
angle, if it falls in that place in the table where the function
used is changing rapidly, the error of observation will
cause a great error in the function to be used.
On the other hand, if, in obtaining the answer to a problem,
it is necessary to look up the angle corresponding to a slowly
changing function, any slight error in the work will cause a
large error in the angle; in other words, if the angle falls
in that part of the table where the function used is changing
so slowly that any slight change in it means a great change
in the angle, a slight error in the function may cause a large
error in the angle. These observations lead to the following:
183b. Rule : If possible, avoid using the tables of functions
for angles between 0° and 5° or 85° and 90°.
Note: The "errors" referred to are not actual mistakes, but errors of ob
servation beyond the Limit of the instrument to detect, while it should be
remembered that the tables are correct only to a certain decimal place. In
the former connection it is well to remember that with an expensive survey
ing instrument it is not easy to measure an angle exactly to the nearest
quarter of a minute. The ordinary school arithmetics, by giving examples in
d3grees, minutes and seconds, are perhaps responsible for the impression that
a second of arc is a quantity easily measured.
Since, in laying out a triangle in surveying, the size of one
GENERAL MATHEMATICS
223
angle affects the size of the others, and it is consequently not
possible to measure all the angles so that they will fall near
the middle of the table, the following rule has been approved
by practice :
183c. Rule : The best form of triangle for computation is
one that approximates the equilateral.
This means practically that it is better not to lay any angle
of the triangle larger than 75° or less than 35°, in order to
obtain the best results in the computation of the angles or
the sides. In laying out a triangle by its sides the same
thing should be kept in mind — that is, the three sides of a
triangle should be approximately of the same length.
(XVI) Measurements for Topographical Map of
Triangulation
Equipment: Transit, ranging pole, measuring tape.
e
Procedure : In taking measurements for the preparation
)f a map of the same field measured in Field Exercise XV
he various prominent features may be located by taking
224 GENERAL MATHEMATICS
two observations of the object with the transit. Thus,
to locate a tree (H) the angles BAH and ABH should be
measured, and since AB is known the length AH and BH
may be solved by the method of the following chapter.
Any other side of the field will do as well as AB for locating
topographical features, since the lengths of these sides are
solvable (Field Exercise XV).
This method serves very well for locating objects of no
great length, horizontally, such as trees, etc., but it will
not do to locate buildings.
To locate the building I J KL, the position of one corner of
the building may be determined by the method just ex
plained (by measuring the angles IBF and BFI). Then a
stake (M ) may be located in the line KJ produced, and
another stake ( N) located in LI produced so that JM = IN.
(The line M N is parallel to the side I J of the building, and
if produced to intersect AB will make the same angle with
A B as IJ would if produced.)
Set up the transit at M and locate the point (0) where
MN produced intersects AB. Set up the instrument atO
and measure BOM. Since the corner N can be located
on the map, the angle BOM will enable us to have the repre
sentation of the building make the proper angles with the
side of the field.
A North and South line may be determined in the follow
ing manner. (The ranging pole will serve for this determi
nation, but it would be better if some much taller object
in the field could be used.) Three hours before noon
mark the shadow of the extremity of the tall object with a
stake. Three hours after noon mark the extremity of the
shadow with a stake. Bisect the angle between the tall
object and the two stakes, and mark the point determined in
the bisector with the stake, P. Produce the line from P to
the tall object so that it intersects one side of. the field. The
angle between this line (which is the North and South line)
GENERAL MATHEMATICS 225
and the side of the field will be the "bearing" of the field
(remember that as we face the north, east is on the right
hand and west on the left).
This determination should be repeated, with reference to
the same side of the field, and the average used.
CHAPTER XVII
SOLUTION OF OBLIQUE TRIANGLES
184. In the solution of oblique triangles, four cases may
be distinguished among the given parts:
(I) One side and any two angles.
(II) Two sides and their included angle.
(III) Three sides.
(IV) Two sides and the angle Opposite one of them.
Notice that in each case three parts are given. One of
the three given parts must be a side: if the three given
parts were the angles, only the shape of the triangle could
be determined — not its size (Chap. XIII). Notice that
in the first three cases the given parts are the same as in the
theorems for proving triangles equal (Chap. IV) — in other
words a triangle is completely determined when these parts
are known.
The first two cases cover nearly all the practical cases oc
curring in surveying with the transit; the third occurs in
surveying with the chain (or tape) alone. The fourth case
is called the ambiguous case and should be avoided in prac
tice.
Case I. Given one side and any two angles.
Note: It is immaterial which two angles are given, since
the third angle may always be f ound by subtracting the sum
of the two given angles from 180° (80a).
The problems occurring under Case I may be solved by
means of the Sine Law, which is derived as follows :
226
GENERAL MATHEMATICS
227
185a. Theorem: In any triangle the sides are proportional
to the sines of their opposite angle.
Part I. When the altitude falls upon the base.
5 a A u
Given: ABC an acuteangled A, in which a, b, and c are
the sides opposite the A A, B, and C, respectively.
To Prove :
a Sin A c Sin C b Sin B
b =
SinB
c
a
Sin A
SinC
Proof : From C draw the altitude CD, or h.
Then, in rt. A CD A, Sin A = r
In rt. A CDB,
SinB = 
a
Dividing the first equation by the second,
Sin A _ h h _ a
SinJ3~6^o~6
In like manner, by drawing the altitude from B, it may
be proved that
SinC _ c
Sin A a
and by drawing the altitude from A it may be proved that
Sin J? b
SinC c
228 GENERAL MATHEMATICS
Let the student complete this part of the proof in del
Part II. When the altitude falls upon the base produ<
Given: Z A is the obtuse Z of the A ABC, in which c
and c are the sides opposite A A, B, and C, respectively.
To Prove :
a _ Sin A
b " Sin 5
*
e = SinC
a Sin A
b Sin J?
c Sin C
Proof : From C draw the altitude CD (or h) meeting •
produced at D.
Then in A BDC,
Sin B = 
a
Z CAD is the supplement of L A,
.. Sin A = Sin CAD (182g).
Now in A CAD,
Sin CAD = J
o
Substituting for Sin CAD its equal, Sin A
Sin A = 
Dividing this last equation by the first,
Sin A _ a
SinB " 6
GENERAL MATHEMATICS 229
As in Part I it may now be proved that
Sin C = c and Sing = b
Sin A a Sin C c
186b. The above equations may be written as proportions,
thus ' a:b = SinArSinB
c:a = Sin C:Sin A
b:c = Sin B: Sin C
186c. By alternation,
a b
Sin A Sin B
c a
Sin C Sin A'
a b
c
(164a)
whence
Sin A Sin B Sin C
(31)
Problem: Solve A ABC, given that Z A = 62° 30',
Z B = 64° 45' and side a = 84.5
Solution: The first step is to form a proportion involving
the three given parts and one unknown part: thus,
I Sin A : Sin B = a : b (185b)
Substituting the values of the given parts
II Sin 62° 30' : Sin 64° 45' = 84.5 : &
60 " 60 " 2 ." 60 ~ 4
Sin 63° = .8910 Sin 65° = .9063
Sin 62° = .8829 Sin 64° = .8988
T.D = .0081 T.D = .0075
$ TD= 0041 3
4 1 .0225
.0056
230 GENERAL MATHEMATICS
Sin 62° = .8829 Sin 64° = .8988
.0041 .0056
Sin 62° 30' = .8870 Sin 64° 45' = .9044
Substituting these values in (II),
.8870 : .9044 = 84.5 :b
.8870 6 = .9044 X 84.5
.9044 X84.5 _ _
.8870
— OU . J.U
To find Z C,
Z A =
62° 30'
179° 60'
Z B =
64° 45'
127° 15'
127° 15'
Z C = 52° 45'
ail) Sin A :
Sin C — a :
c.
(IV) Sin 62°
30' : Sin 52°
45'
= 84.5 :c
45' =
3°
4
•
Sin 53° =
.7986 •
Sin 52° = .7880
Sin 52° =
.7880
.0106
.0080
Sin 52° 45' = .7960
3
n
.0318
.0080
Sin 62° 30' has already been found.
Substituting these values in (IV),
.8870 : .7960 = 84.5 :c
.8870c = .7960 X 84.5
_ .7960 X 84.5
C ~ 78870 75  83
GENERAL MATHEMATICS 231
Notice that in solving for c, the Z A and side a are used,
as in solving for 6. This is the best method — to use the given
parts in solving for both of the unknown sides. If b had
been used in solving for c, an error (due to the inaccuracy of
the decimals) occurring in b would be increased in carrying
it through the computation for c.
Notice the advantage by this method of computation of
having the angles of nearly the same size — that is, of having
a triangle that is nearly equilateral. In the above com
putation,
.7960 X 84.5
C "~ .8870
Since the sine of Z A (.8870) is not very greatly different
in value from the sine of Z C (.7960), in the above fraction
84 . 5 is approximately multiplied and divided by the same
quantity — that is, any error of computation or measurement
in the terms of the above fraction is not increased as it would
be if the sine in the denominator were much smaller than
the sine in the numerator.
Exercise
Solve the triangle ABC. given that:
1.
A = 61° 20',
B = 62°,
b = 101
2.
A = 59° 10',
C = 64° 50',
c = 95
3.
B = 56° 40',
C = 63° 25',
a = 201
4.
A = 41° 48',
B = 72° 35',
b = 97
5.
C = 47° 36',
B = 65° 35',
c = 117
8.
C = 43° 24',
B = 71° 6',
a = 1.46
7.
A = 58° 30',
C = 59° 45',
b = 2.56
Case II. Given two sides and their included angie.
Problems occurring under Case II may be solved by aid
of the following theorem, which is known as the Cosine Law.
232 GENERAL MATHEMATICS
186a. Theorem : In any oblique triangle the square of any
side is equal to the sum of the squares of the other two sides minus
twice their product times the cosine of their included angle.
Part I. When the altitude falls upon the base.
T D
Given: ABC an acuteangled triangle, in which a, b, and c
are the sides opposite the A A, B } and C, respectively.
To Prove :
a a = 52 + C 2 _ 2bc Cos A
b* = a* + c 2  2ac Cos B
c* = a 2 + b*  2ab Cos C
Proof: From B let fall the perpendicular h } intersecting
AC at D.
Then AD is the projection of c on AC (149).
I. .\ a 2 = & 2 + c 2  26 (AD) (150b).
But in rt. A ABD,
c
Multiplying both sides of this equation by c,
AD = c Cos A
Substituting this value for AD in (I),
a 2 = 52 + C 2 _ 2bc Cos A.
In like manner, by drawing the altitude from C, it may be
proved that
& 2 = a 2 + c 2  2ac Cos B
Also by drawing the altitude from A it may be proved
that
c % = &2 + a 2 _ 2a& Cos C
GENERAL MATHEMATICS 233
Let the student complete this part of the proof in detail.
Part II. When the altitude falls upon the base produced.
Given: Z A is the obtuse Z of the A ABC, in which
0> by and c are the sides opposite 4 A, B, and C, respectively.
a* = b 2 + c 2  2bc Cos A
62 = a % + C 2 _ 2ac Cos B
c* = a 2 + ft 2  2a& Cos C
Proof : From B draw the altitude fc, meeting C A produced
at D.
Then DA is the projection of c on & (149).
II. .. a 2 = & 2 + c 2 + 26 (AD) . (150c)
Since DC is a straight line, Z BAD is the supplement of
Z £AC (43b).
/. Cos Z DAB =  Cos Z BAC (182g).
Now in A BAD,
A D
Cos Z DAB = —
c
Whence, multiplying both sides of the equation by c,
DA = c Cos Z DAB.
Substituting ( Cos BAC) for its equal Cos Z DAB in this
equation, DA = c (  Cos Z BAC).
Substituting this value for DA in (II),
a* = &* + c 2 + 2&c (  Cos Z BAC)
Whence,
a 2 = 52 + C 2 _ 2&c (Cos Z BAC)
234 GENERAL MATHEMATICS
Since in the given triangle (exclusive of construction
lines) there is only one angle at A, Z BAC may be written
Z A and so the above equation becomes
a 2 = b 2 + c*  26c Cos A
It may now be proved as before that
62 = a 2 + C 2 _ 2ac Cos B
c* = a 2 + 6 2  2a6 Cos C Q. & D.
If Z A is a right angle, cos A is 0, and so the factor 26c
Cos A becomes (since times any quantity is 0) ; that is,
if Z A is 90° the equation
a 2 = 6 2 + c*  26c Cos A
reduces to
a 2 = 6 2 + c 2
Since here a represents the side opposite the right angle,
the formula therefore is a statement of the Pythagorean
Theorem.
In the same manner it may be proved that
6 2 = a 2 + c 2  2ac Cos B
C 2 = 6 2 + a 2  2ab Cos C
reduces to the Pythagorean Theorem.
Problem: Solve for the remaining parts of the triangle
ABC, given that a = 48, 6 = 52, C = 58° 10'.
Solution: c 2 = a 2 + 6 2  2ab Cos C (186a).
Extracting the square root of both sides of the equation,
c = Va 2 + 6 2  2a6 Cos C
Substituting the values of the given parts^
c = V(48) 2 + (52) 2  2 (48) (52) (.5275)
= V2304 + 2704  (4992) (.5275)
= V5008 2633.28
= V2374.72
= 48.73
GENERAL MATHEMATICS 235
Z A may now be found by aid of the Sine Law, thus
Sin A: a = Sin C: c
Substituting the value of aine of Z C (58° 10') from the .
table and also substituting the values of a and of c,
Sin A : 48 = .8496 : 48.73
(48.73) (Sin A) = 48 (.8496)
48 ( . 8496)
SmA = ~4&7is • 8368
Z A = 56° 48'
Z B might now be found by adding Z A and Z C and
then subtracting their sum from 180, but it is better to find
Z B independantly, and then use the theorem that the sum
of the three angles is 180 as a check on the work, thus:
Sin B : b = Sin C : c
Sin B : 52 = .8496 : 48.73
(48.73) (SinB) = (.8496) (52)
gin B = C84961J52) = 9Q66
48. 16
L B = 65° 2' Check
Z A = 56° 48'
Z B = 65° 2'
Z C = 58° 10 ;
180°
186b. For use in solving such problems as the above, the
formulas of 186a may be modified as follows:
a = Vb 2 + c 2  26c cos A
b = Va 2 + c 2  2ac cos B
c = Va 2 + b 2  2ab cos C
236 GENERAL MATHEMATICS
Exercise
Solve the A ABC, given that:
1. a  432, 6 = 321, C = 57° 30'.
2. c = 371, & = 325, A = 62° 24'.
3. b = 231, a = 263, C = 59° 30'.
4. c = 432, a  394, A = 49° 35'.
5. & = 243, c = 262, A = 70° 15'.
Case III. Given the three sides.
By means of the Cosine Law the angles of a triangle may
be computed when the three sides are known. By 186a,
a* = b* + c*  26c Cos A
Transposing,
26c Cos A = 6* + c 1  a*
Dividing by the coefficient of Cos A,
^ . 6* + c s  a»
C ° SA 26^
In like manner the value of Cos B may be found from the
equation,
6 2 = a 2 + c 2  2ac Cos B
and the value of Cos C from the equation,
C 2 = a 2 + b 2  2ab Cos C
Hence,
186c.
~ A b 2 + c 2  a 2
C0SA= Wc
a 2 + c 2  ¥
CosB =
CosC =
2ac
a 2 +b 2  c 2
2ab
These formulas may all be included in the following:
GENERAL MATHEMATICS 237
Rule : In any triangle the cosine of any angle is equal to the
sum of the squares of the two sides including the angle, minus
the square of the opposite side, all divided by twice the product
of the two sides including the angle.
Problem: Solve for the angles of the A ABC, given that
a = 232, 6 = 325, and c = 251.
Solution: First square all three sides:
a* = 53,824, ¥ = 105,625, c* = 63,001
Next substitute the given values in the formulas (186c).
Do not substitute in the first formula only, and then work
that out, then substitute in the second, and work that out,
etc., but substitute in all three. This is much less confusing.
Thus:
105,625 +'63,001  53,824
Cos A =
2325251
A =
Cog B = 53,824 + 63,001  105,625
2232251
Cog c = 53,824 + 105,625  63,001
2232325
B
C =
Check :
This arrangement of the work (called a " skeleton") has
the advantage of separating the more exacting work of sub
stituting in the formula from the more mechanical work of
computation. The values of the cosines may now be com
puted. After all three of the cosines have been computed, the
table of sines and cosines may be opened and the values of
all three cosines written down at the same time.
Remember that the rule for rapid and accurate computa
tion is "never open the table of functions until everything
possible has been done without it."
238 GENERAL MATHEMATICS
The work now appears as follows:
, A 105,625 + 63,001  53,824
CoS A 2325251  7037
A = 45° 17'
53,824 + 63,001  105,625
C ° S B 2232251 =  0962
Z B = 84° 29'
53,824 + 105,625  63,001
C ° S C 2232325  6396
L C = 50° 15'
O 1 /"
Check, 180° 1
• The three angles should always be computed separately
i^as in the above arrangement) and added to test the accuracy
of the work. To compute two of the angles and subtract
their 3um from 180 would "lump" the error of computation
on one angle.
Exercise
Find the angles of the A ABC, given that
1. a = 27, b = 32, c = 29.
2. a = 52, b = 56, c = 48.
3. a = 107, b = 112, c = 121.
4. a = 2.34, b = 2.56, c = 2.24.
6. a = 982, 6 = 1,003, c = 1,175.
6. a = 5,218, 6 = 4,927, c = 5,127.
186d. It must be borne in mind that in practical work, all
measurements are subject to errors of observation. For
example, in measuring one side of a triangle errors may arise
due to the stretching of the tape, the irregularity of the
ground, and other causes. If, in measuring one side of a
triangle sufficient care was taken (by leveling the tape, etc.,
and taking the mean of repeated observations) so that the
result was correct to the nearest tenth of a foot, it would be
GENERAL MATHEMATICS 239
absurd to combine this in computation with a side which was
accurate only to the nearest foot — as the computed side
would be correct only to the nearest foot.
The number of figures in the length of a side is considered
as indicating the accuracy with which the measurements
have been made. Thus, if one side of a rectangle is 8 feet
long and another 79 feet long, the presumption is that each
side is correct only to the nearest foot, but as one foot is a
much larger percentage of 8 feet than of 79 feet, the longer
side would be considered as being more accurately measured,
and to combine these two sides in computation would disre
gard the principle laid down in the preceding paragraph.
To obviate this, the 8 foot side should be measured to the
nearest tenth of a foot. Note that if . 3 of a foot were dis
regarded in the short side of this rectangle the area would be
in error by nearly 24 sq. ft. Hence,
Rule : Let all lines, both measured and calculated, show the
same number of figures.
Note: This rule may be disregarded if the sides are nearly equal in value;
for example, a side 98 feet long may be combined in computation with one
103 feet long.
Obviously, the computed results can be no more accurate
thail the measurements — it is useless to carry out the com
putation of a side to more places of decimals than a meas
ured side. The same principles apply to the angles.
The degree of precision in the computations, consistent
with the accuracy of the data, is as follows:
187. Two significant figures in the sides, angles correct to
the nearest half degree.
Three significant figures in the sides, angles correct to
the nearest 5 '.
Four significant figures in the sides, angles correct to the
nearest minute.
The following theorems will also be of use in checking the
results:
240 GENERAL MATHEMATICS
188a. // the sides of a triangle are unequal, the largest angle
is opposite the longest side, and the smallest angle is opposite
the shortest side.
Given: In A ABC, BC is the longest side and AB is the
shortest side.
To Prove : Z A is the largest Z and Z C is the smallest Z .
(Z B means Z ABC, and Z A means Z BAC.)
Proof: Lay off AD = AB.
Draw BD.
(Z ADB is the exterior Z of the A BBC, and also one
of the equal angles of the isosceles A ABD.)
Z ADB > Z C (81c)
Z ABD = Z AD5 (77b).
Substituting in the above inequality,
(I) Z ABD > AC.
(II) Z B > Z A52) (33a).
Much more so, then,
(III) Z B> AC.
The reasoning by which (I) and (II) are combined to obtain (III) is much
the same as saying, "5 > 3, 7 > 5, much more so, then, is 7 > 3."
Lay off EC = AC.
Draw AE.
GENERAL MATHEMATICS 241
(Z AEC is the ext. Z of the A AEB, and also one of the
equal angles of the A CAE.)
Z AEC > LB (81c).
But Z AEC = Z EAC (77b).
Substituting for Z AJ?C its value in the above inequality,
(IV) Z EAC > Z B.
(V) But Z A > Z JSAC.
Much more so, then, is
(VI) Z A > Z B.
By (VI) and (III), since Z A> Z B and Z B > Z C.
(VII) Z A> Z C.
By (VI) and (VII),
Z A is the largest Z of the A.
(Since it is greater than either of the others.) The small
est angle must be one of the remaining two, and since Z B
> Z C, Z C is the smallest. Q. E. D.
If the results of any problem are inconsistent with this
theorem, the work is wrong. Let the student apply tliis
test to the answers to the problem of 186c.
188b. Theorem : // the angles of a triangle are unequal, the
longest side is opposite the largest angle, and the shortest side
is opposite the smallest angle.
Given: In A ABC, Z A is the largest Z and Z C is the
smallest Z.
To Prove: BC is the longest side and A B is the shortest
side.
242 GENERAL MATHEMATICS
Proof: The relation between BC and AC must be one of
these three:
(1) BC = AC.
(2) BC < AC.
(3) BC > AC.
(1) BC cannot be equal to AC, for if it were A ABC
would be isosceles and Z A would equal Z B (77b).
But this contradicts what is given, that Z A is the
largest Z.
.\ (1) cannot be true.
(2) BC cannot be less than AC, for if it were LB
would be larger than Z A (188a).
But this contradicts what is given, that Z A is the
largest Z.
/. (2) cannot be true.
/. (3) the only remaining relation must be true, that is,
a) BC > AC.
, The relation between AC and A B must be one of these
three:
(4) AC = AB.
(5) AC < AB.
(6) AC> AB.
(4) AC cannot be equal to AB, for if it were* then A
ABC would be isosceles, and Z B would equal Z C.
But this contradicts what is given, that Z C is the
smallest Z.
.*. (4) cannot be true.
(5) AC cannot be less than AB, for if it were Z C
would not be the smallest Z .
But this contradicts what is given, that Z C is the
smallest Z.
.\ (5) cannot be true, that is,
(II) AC > AB.
By (I) and (II),
BC > AC, and AC > AB.
GENERAL MATHEMATICS 243
Hence,
(III) BC > AB.
By (I) and (III),
BC is the longest side of the A.
(Since it is longer than either of the others.)
The shortest side must be one of the two remaining sides,
and since AC > AB, AB is the shortest side. Q. E. D.
If the results of any problem are inconsistent with this
theorem the work is wrong.
Let the student apply this to the answers to the problem
under the Sine Law. (Page 230.)
189. Theorem: The ratio of any side of a triangle to the
sine of the opposite angle is numerically equal to the diameter
of (he circumscribed circle.
Given: is the center of the circle circumscribed about
A ABC, and R its radius; a, b, and c the sides opposite the
A A, B, and C, respectively.
To Prove: 2*^^^
Proof: Draw OB and OC; draw OF _L to BC.
£ BOC is an isos. A (18).
OF bisects Z BOC (80d).
244 GENERAL MATHEMATICS
That is,
(I) Z BOF = 1/2 Z BOO.
Again, Z A is measured by 1/2 arc BC (89a).
Z BOC is measured by all the arc BC (26).
(II) .. Zi = 1/2 Z £OC.
By (I) and (II),
(III) Z BOF = Z A.
BF = 1/2 o (80d).
That is,
(IV) a = 2BF.
In the rt. A BOF,
BF = R Sin Z BOF (172)
= # Sin A (by III)
By (IV)
a = 2R Sin A.
Dividing both sides of the equation by Sin A,
®^A =2R
By the Sine Law,
a b c
Sin A Sin B Sin C
Whence,
a b
222 =
Sin A Sin B Sin C QE.D.
Exercise
l. Let the student work out another proof for the Sine
Law, as follows : As in the above proof, from A BOC prove
that 57^r = 222.
Sin A
In like manner, by drawing 0A, and a _L from to 6, it
may be proved that ~ — ~ = 222.
GENERAL MATHEMATICS 245
Also, by drawing a J_ from to c prove that ^ — ~ = 2R.
Equate these three values of 222.
2. Draw a triangle and circumscribe a circle about it.
Measure the sides of the triangle and the diameter of the
circle : change all fractions of an inch to decimals. Measure
the angles with a protractor, look up their sines, and divide
each side by the sine of the opposite angle. Compare the
quotients thus obtained with the diameter of the circle.
3. The diagonal of a parallelogram is 8 centimeters long,
the adjacent sides at one of its extremities make angles of
43° 17' and 75° 42' with it. Find the lengths of the sides
(64, 122b).
4. An observer knows that he is 2 . 7 miles from one fort
and 3.6 miles from another; from where he stands the dis
tance between the forts subtends an angle of 24° 29'. Find
the distance between the forts.
6. A roof in which the rafters have a rise of 9 ft. and a run
of 13 ft. is to be supported by a "scissorbeam" truss con
structed according to the following diagram:
The middle of each rafter is to be supported by a strut
reaching to the middle point (D) of a beam stretching from
A to C. Find the distance from the middle points of the
rafters (E and F) to D.
Hint: First solve rt. A ABD and obtain length A B and
ZABD. Solve A EBD in which sides EB (= ~\
and BD, and Z EBD are now known.
J
246 GENERAL MATHEMATICS
6. Starting from a point A, a surveyor runs a line due north
40 chains to B, he then runs north 64° west 40 chains to C.
How far is C from At What is the bearing of C from At
7. At the base of a hill the angle of elevation of the
summit is 9° 50'; 3000 feet farther up the hill (along a slope
which is inclined at $n angle of 7° 30' to the horizontal) the
angle of elevation is found to be 15° 45'. Find the height of
the hill.
8. An isosceles triangle is inscribed in a circle, the radius
of which is 5 in. ; the base of the triangle is 5 . 5 in. Find
the vertical angle (the angle opposite the base) and length
of the other sides.
Hint: Draw a diagram; draw the radii to the three ver
tices.
9. A roof in which the rafters have a run of 19 feet and a
rise of 10 feet is to be supported by a scissorbeam truss con
structed according to the following diagram:
The middle point of each rafter (D and E) is to be sup
ported by a strut reaching to the opposite plates (A and C),
respectively. Find the length of DC and EA.
Hint: Find Z B by solving the rt. A AGB for Z ABG
and doubling it (80d).
Find the bevel at which the upper ends of the struts are
to be cut, that is, Z ADC, also the lower ends, Z DC A.
Note : It is not necessary to make any allowance for the
plate cut, as that may be laid out afterwards, in the same
manner as for the rafter (Shop Exercise XI Va).
Find also the distance BF.
Hint: In A BDF } Z BDF (supplement of Z ADC) and
Z DBF and DB are now known.
GENERAL MATHEMATICS
247
10. Two sides of a parallelogram are 3 in. and 4 in., respec
tively, and the angle between them is 41°. Find the length
of the two diagonals.
A formula for finding the area of a triangle when its three
sides are given has already been derived (155). The actual
labor of working out the area, however, may usually be
much diminished by the following:
190a. Theorem: The area of any triangle equals half the
product of any two sides times the sine of their included angle.
Given: a, 6, and c the sides opposite the A A, 23/ and C,
respectively, of the A ABC; h the altitude from vertex A to
base a.
To Prove:
a i t^^ ac Sin B ab Sin C be Sin A
Area A ABC = zz = = —
Proof: Area A ABC = 2£* (133a).
But h = c Sin B (172). oc Sin 5
Substituting in the above, Area A ABC = ~
In like manner, by drawing altitudes from the other ver
tices, let the student prove that
a nsv ab Sin C
Area A ABC = 
Area A ABC =
2
be Sin A
Note that the above theorem applies equally well to both
figures (either the acuteangled or obtuseangled triangle).
248 GENERAL MATHEMATICS
This fact is of service in fixing the above theorem in the
memory, since it indicates that the function used is the
sine — as, if it were the cosine, there would be two theorems,
since the cosine of an obtuse angle is negative (182g).
The solutions of triangles (such as in the preceding chap
ter) may be checked by finding the area with regard to two
different angles; for example, by formula — > and then
by These results should check, within the error
allowable through use of approximate decimals.
Exercise
1. Find the area of the triangles in ex. 6 and ex. 8 in the
above exercise.
2. Check the results of the problems under the Sine Law
by finding the area of each triangle, by means of the angle
and its two including sides, at two different vertices.
Example: The problem worked out under the Sine Law
was as follows :
Given: A = 62° 30', B = 64° 45', a = 84.5.
Computed: C = 52° 45', b = 86.16, c = 75.83.
Check: ^  86,16 X 7588 X. 8870 . 2897.61
acSmB 84.50 X 75.83 X .9044 rto ^ M
— ^ — = ~  2897.53
XVII. Drawing Exercise
MAP OF SURVEY
Perform all necessary computations to complete the work
of Field Exercises XV and XVI.
Keeping in mind that the top of a map should be approxi
mately North, make a careful drawing to scale, locating the
GENERAL MATHEMATICS
249
various topographical features by the aid of the measure
ments made in Field Exercise XVI, or computed from them.
In the upper left hand corner, within the marginal line en
closing the map, draw a line running exactly North and
South ; mark its upper end with an arrowhead aiid above it
an "N," the lower end with "S." By drawing light pencil
lines parallel to this, through the proper points in the draw
ing, the bearings of the fences/etc, measured in the previous
Field Exercise may be laid off with the protractor.
■ VI
■fVS
f 1 *
\T~ .— .»»«•. _««•. ..«.•,.
.1— .— «•». . t?.
■  —   —  _•— —•^— ^««» ..^»— ..»•■— .._••*
The accompanying diagram shows various conventional
topographical signs; upper right hand corner, cornfield, and
below it ploughed land; upper left hand corner, orchard;
lower left hand corner grassland. The brook, fences,
buildings and grounds can easily be identified in the drawing.
The area included by the map (found by summing the
areas of the triangles measured in Field Exercise XV)
should be marked on the map in acres. To complete the
map, its title (in large letters) and the scale used (for example
"1' = 1 chain"), the name of the observer, and the date (in
smaller letters) should be arranged neatly in the lower right
hand corner within the marginal line enclosing the map.
CHAPTER XVIII
SIMULTANEOUS EQUATIONS: FINDING HEIGHTS
OF HILLS: AMBIGUOUS CASE*
191a. Problems involving two unknowns may be solved
by two equations containing both x and y, which must be
solved together, and hence are called simultaneous equa
tions. Thus,
Problem : A rectangle is 5 feet longer than it is wide and
its perimeter is 50 feet. Find its length and width.
Solution: Let x = length
y =* width
Then
(I) x  y = 5
(II) 2x + 2y = 50
Multiplying (I) by 2, and adding (II) to it
2x 2y = 10
2x + 2y = 50
4x =60
x = 15
Substituting this value in (I)
15  y = 5
Transposing
y= 10
Changing signs throughout (dividing by — 1)
y = 10
Check: Substituting the values of x and y in (II)
30 + 20 = 50
Equation I is the algebraic statement that the length of the rectangle is 5
feet more than its width. Equation II is obtained by collecting terms in the
equation which states that the perimeter is 50 feet: x\y + x\y = 5Q.
250
GENERAL MATHEMATICS 251
Solve the following systems of simultaneous equations:
l. 2x + 4y = 16 2. 4x  y = 11
7x + y = 17 2x + 6y = 64
In 3, after multiplying by the proper factors, we may
either add the y terms or subtract the x terms.
3. 3x  y = 15 4. 5x + y = 43
2x + 3?/ = 32 2x + 7y = 37
6. 9x + y = 19
20t  y = 10
191b. Any method of getting rid of an unknown quantity
is called a process of elimination.
The following method of elimination is called substitution.
I. x  y = 5
II. 2x + 2y = 50
From (I): x = y + 5
Substituting in (II)
2 (y + 5) + 2y = 50
2y + 10 + 2y = 50
4y = 40
2/ = 10
Substituting this value in
x = y + 5
s =. io + 5
x = 15
Solve the following sets of simultaneous equations by sub
stitution :
6. 7x + 2y = 39 10. Alx + .02y = 1.24
3x  2y = 11 .92x + .0% = 2.26
7. 3z + 2y = 26 11. .108z + .0212/ = .232
9z + 5y = 71 .107x  .063y = .166
8. 5z  y = 1 12. a; + 6y = 23.4
3z + 2y = 28 hx  3y = 21 .3
9. .4z + .Sy = 4.1 13. 7x + 3y = 4.651
x + .4y = 7.8 4x + y = 1.722
252 GENERAL MATHEMATICS
191c. From the above examples it is evident that elimi
nation by substitution consists in finding the value of one of
the unknowns from one equation and substituting this value
for the unknown in the other equation.
19 Id. Another form of this process is to find the value of
one unknown in terms of the other from one equation; then
find .the value of the same unknown from the other equation,
and equate these two values.
Thus, the above problem may be solved in this manner:
(I) x  y = 5
(II) 2x + 2y = 50
From (I) by transposition
(III) x = y + 5
From (II) by transposition
2x = 50  2y whence
(IV) x = 25  y
Equating the values of x in (III) and (IV)
y + 5 = 25  y
Transposing
2y = 20
y = 10
Substituting in (I)
x  10 = 5
x = 15
Solve the following sets of simultaneous equations by
equating the values of one unknown in terms of the other:
14. 6x2y = 26 17. .5x.3y = .0273
4x y = 22 l.lx.2y=.1842
15. .3x+1.2y=.3534 18. .3xl.ly=  3.4
.4x+ .5y=.436 .6x+ y= 9.2
16. .llx+.03y=1.20 • 19. .112x+.031y=.398
.02x + .ll2/= .95 .108x+057j/=.438
GENERAL MATHEMATICS
253
192. By aid of simultaneous equations it is possible to find
the height of objects when they could not be found by the
method .already learned (Chap. XIV). By that method it
was necessary to measure the base of a right triangle and the
angle of elevation. Sometimes it is not possible to measure
this base line, as, for example, in finding the height of a hill.
In attempting to find BC, the height of the hill, it is ob
viously impossible to measure the part EC of the base line,
which is within the hill. In this case, BC 1 may be found by
the following method:
Suppose the angle of elevation A to have been measured
and to have been found to be 19°, the angle of elevation at D
(in a straight line toward the hill) to be 24°, and the distance
AD to be 300 feet.
Let DC be represented by x, and BC by y.
From rt. A BCA,
Tan A = Tn = a r\ ■ —
AC AD + x
Clearing of fractions
(I) (AD + x) (Tan A) = y
From rt. A BCD,
Tan Z BDC = 
x
Clearing of fractions
(II) x tan Z BDC = y
254 GENERAL MATHEMATICS
Equating these two values of y
(AD + x) (tan A) = x tan Z BDC
Substituting the numerical values for AD, tan 19° and
tan 24°,
(300 + x) (.3443) = z(.4452)
103.29 + .3443* = .4452a;
103.29 = .1009a;
x = 1023.6
Substituting in (II)
y = (1023.6) (.4452) = 456 ft. Ans.
Note that A and D are in the same straight line with the
hill and on the same level. For smaller elevations (or when
ever greater accuracy is desired) a correction should be made
for the height of the instrument — that is, A and D are not
really on the same level as the base of the hill, but the height
of the instrument above it, and this height should be added
to the computed height of the hill (AA' = DD' = CC).
Exercise
1. From the first of two stations on a level road leading
directly toward a hill, the angle of elevation of the top of the
hill is observed to be 12°. From the second station, 200 feet
nearer the hill, the angle of elevation is observed to be 18° 30'.
Find the height of the hill.
2. From a station on a level plain a military engineer
observes the angle of elevation of the top of a hostile fort
to be 4° 15'. From a second station 320 feet directly
toward the fort, the angle of elevation is 6° 20'. Find the
height of the fort.
3. From the bank of a river, the angle of elevation of the
top of a tree on the other bank is observed to be 32° 16'.
From a station 100 feet back of the first, and in the same
straight line with the tree and the first station, the angle of
elevation to the top of the tree is 15° 45'. Find the width of
the river.
GENERAL MATHEMATICS
255
Case IV in the solution of triangles (see art. 184) should be
avoided in practice whenever possible. It is called the
Ambiguous Case, as oftentimes there are two triangles
which satisfy the conditions of the problem, and in case of
a practical application it is necessary to determine which to
use from other considerations. This ambiguity arises from
the fact that the sine of an angle is equal to the sine of its
supplement (Chap. XVI). (In the problems previously
worked out under the Sine Law, this caused no uncertainty,
as in those problems two angles were given and the third
was determined as the supplement of the sum of the two given
angles. )
193a. The classes of examples occurring under the Am
biguous Case may he divided as follows:
I. Impossible.
. II. Two Solutions.
III. One Solution.
Consider the given parts as a, b and A of the A ABC.
If a < b y and Z A is an obtuse angle, Z B must be obtuse
(188a). But this is impossible (147b).
In like manner, the prpblem is impossible if a = b.
In like manner, if A is a rt. Z and a < b or a = b the
triangle is impossible.
Let the student demonstrate this graphically by attempt
ing to draw triangles under the above conditions, when it will
become evident that such triangles will not "close." Hence,
193b. Rule : If the given angle is a right angle or greater,
the side opposite must be greater than the other given side, or
the triangle is impossible.
256
GENERAL MATHEMATICS
Consider first that the given parts a, b ana A are parts of
a right triangle.
Then a = b Sin A (172).
If, however, the value of a is less than b Sin A, evidently
a is not long enough to reach from C to B — that is, the tri
angle would not close. Hence,
193c. Rule : If the given angle is acute, and the side opposite
is less than the product of the sine of the given angle and the
other given side, the triangle is impossible.
To investigate the remaining cases, suppose that a were
gradually increased. It might then cut the line AB to the
left or to the right of the J_ CB, as in the following figures.
In the first figure a falls within BC, in the second it falls
without. Both these triangles fulfill the given conditions
having the given sides a and b, and the given angle A.
Consider now that the side a in the first figure is gradually
lengthened; B', its point of intersection with AB, comes
nearer and nearer to A. When a = b, the two lines coincide,
and there is no triangle. If a continues to lengthen, B' will
fall outside of A, and the angle at A will no longer be the
given angle, but its supplement — and the left hand figure no
longer fulfills the given conditions. In this figure, then, the
value of a must lie between b Sin A and b.
In the right hand figure, however, a may evidently be
increased indefinitely (the point B 1 moving continually to
GENERAL MATHEMATICS
257
the right) without violating the conditions of the given
triangle.
Since in both figures the triangles begin with a = b Sin A
(when a < b the triangles do not close) and in the left hand
figure the triangles cease when a = b, while in the right hand
figure they continue indefinitely, it follows:
193d. Rule : If the given angle is acute, and the side opposite
the given angle is greater than the product of the other given side
times the sine of the given angle 9 , but less than the other given side,
there are two solutions.
193e. Rule : If the given angle is acute, and the side opposite
the given angle is greater than the other given side, there is one
solution.
The following figure is a graphic representation of these
rules.
<f
i
I
I
i
4
I
I
M
4
%
s
s
Ott* J+h>f*tr
•
Zon%
9
*
•
75w &ol*tior*
Ze/r«
• *
• •
• ■
* s
i i
1*
No 3oiuti**9
2o/r«
»
•
•
: i
•
•
\
% i
•
I :
i
* i «
i i
■ *
i • •
A
6
XVIII. Field Exercise
Find the height of a hill by the method outlined in this
chapter.